I don't have enough time now to write a good diary entry -- I'm sorry!

The Riemann sphere and stereographic projection
One geometric method which helps understand certain conformal mappings is to visualize what is going on using stereographic projection. Put the complex plane into space (R³) as the xy-plane, with third coordinate equal to 0. Draw a sphere with radius 1 and center the origin, (0,0,0). Let N, the North Pole, be located at (0,0,1). Then every point a on the sphere which isn't N corresponds to a unique point b on the complex plane. The correspondance is by drawing a line from N to a and extending it until it intersects the plane. The intersection point is called b. Of course, we could start with b on the complex plane, and connect it with N, and the line would always hit the sphere in some other point, a. This "correspondance" is a useful thing to keep in the back of your mind (?) throughout your study of the subject. Of course you can write lots and lots of equations describing everything, but right now I just want the idea, without proofs or whatever. In fact, it turns out that stereographic project also "preserves angles" between corresponding curves (this is not immediately obvious at all).

Comment The stereographic project shown here is what's usually done now. But there is another common variant, where the sphere is "balanced" on top of the plane, so its center is at (0,0,1/2) and the North Pole is at (0,0,1). This is what's done in the video by Arnold and Rogness.

Now back to D₁(0) and H
We then struggled to see what happens to the unit disc. It corresponds under stereographic projection to the lower half of the sphere. We could then rotate the sphere and make the lower half into the "back" half. This then corresponds to H, the upper half plane. So: we see that D₁(0) and H are the same. We had previously shown that z⇒(z–i)/(z+i) maps H to D₁ in a 1-1, onto, conformal fashion.

Linear fractional transformations

And 2-by-2 matrices Triple transitivity

Circles and lines

Examples
From a past exam and from class examples.

Diangles

1. The conformal mapping property of analytic functions (why is this interesting?)
2. Linear fractional transformations and the Riemann sphere.

Here are appropriate textbook homework problems for some of the content of the last few meetings of the course:

3.1: 10, 11, 12, 14, 18, 20; 3.3 4 a), b), d); 5 a), b), c); 7a), b), d).

Linear algebra and angle preservation
Now I hope to do a linear algebra exercise. I will find all linear maps from R² to R² which preserve angles. Examples of such mappings are rotations, dilations (just multiplying lengths), and reflections (such as complex conjugation -- reflection across the real axis).
I will do this with an informal discussion -- a sequence of observations. I will suppose that the linear map has matrix M given by

( a b ) 
( c d )

1. What if det(M)=0?
2. More to come ...

( 1 b )
( 0 d )

( a b )
(–b a )

( 1 0 )
( 0 –1)

[Direct] conformality and Cauchy-Riemann
A differentiable mapping from R² to R² is called (very precisely!) directly conformal if the angles between two differentiable curves meeting in the domain are preserved in the range and if the orientation is preserved. Most people omit the word "directly" and assume that orientation is preserved. Now we will observe that direct conformality implies that the matrix of first partials of the mapping must satisfy the Cauchy-Riemann equations. So we have a geometric equivalence for all the other characterizations of analyticity.
Well, not exactly: the matrix

( a b )
(–b a )

Local characterizations of analytic mappings
If f´(z₀)=a+i b≠0, then locally f multiplies tangent vectors of curves by a+i b: it rotates by Arg(f´(z₀) and it magnifies by |f´(z₀)|. But what if the derivative is 0?

Examples of this are z² and z³. Consider them near 0. These are not 1-1 there (except at 0). They are locally 2-1 and 3-1 respectively. The angular behavior is interesting: there's a multiplication of central angles by factors of 2 and 3 respectively. It truns out that this is representative of what happens in general.

Suppose that f(z)=∑_n=0^∞a_n(z–p)ⁿ, where this sum is valid near p and f is not the constant function. Suppose now f has a zero of order N>0 at p. This means that a bunch of derivatives (N of them) are 0. So the coefficients a₀, a₁, a₂, ..., a_N-1 are all 0. Then f(z)=(z–p)^Nh(z), where h(z)=a_N+a_N+1(z–p)¹+a_N+2(z–p)² +...=∑_n=N^∞a_n(z–p)^n-N. Now h(p)=a_N and (most important for this discussion because the non-zeroness gives conformality) h(p)=a_N≠0. Then we know that h(z) is analytic and not 0 in some disc around p. So we saw two lectures ago that h(z) has a logarithm: we can write h(z)=e^k(z) for some analytic k(z), valid near p. But define j(z)=e^k(z)/N. Then j(z)^N=h(z) near p, so that f(z)=((z–p)j(z))^N. This means that locally near p I can think of f(z) as a composition of (z–p)j(z) which has a non-zero derivative at p (this comes from a_N≠0) and the mapping z^N. So all (non-constant) analytic functions are locally compositions of simple powers (if the derivative is 0) and conformal mappings.

I believe this discussion (copied, with some changes, from a text by Ahlfors) is now correct. It differs in some details with what I did in class. Mr. Marcus pointed out some errors (or, more diplomatically, some confusion!) to me after class.

One consequence
I will merely mention this: all non-constant analytic functions are open: the image of any open set is an open set. This is weird and is certainly not true for differentiable functions on R to R (x⇒x²) or from R² to R (consider (x,y)⇒x²+y²) or from R² to R² (consider (x,y)⇒(x²,y) -- "folding" over the y-axis). This openness has itself strong consequences (the Maximum Modulus Theorem). So the action of folding a piece of paper (as I mentioned in class) can't be analytic.

A BIG theorem
Consider any connected, simply connected (no holes!) open set in C. If this set is not the whole plane, then there is a conformal mapping from the set to the inside of the unit disc.

This is called the Riemann Mapping Theorem. We can't include all of C because the result would be false (a consequence of Liouville). But everything else has a really nice, simple model. I tried to show an example of a simply connected open set (C with a logarithmic spiral removed) to exhibit something for which the theorem is not at all obvious. Indeed, the theorem's conclusion can hardly be believed for this open set.

There are many proofs of this result. Riemann's own idea of a proof was flawed: see the discussion here, for example. It depended on being able to solve the Dirichlet problem. Now most people would actually assert that proving the RMT and solving the Dirichlet problem (at leaast in 2 dimensions) are essentially the same. In this course, with hardly any time left, I'd like to show shown examples of mappings which verify RMT. We already know some but ...

We will get the beginnings of a very small dictionary of conformal mappings. We know because of (z–i)/(z+i) that the upper half plane is the "same" as the unit disc. By using powers (where the non-conformality is on the boundary!) we can conformally map wedges (such as the first quadrant) to the unit disc. By using the exponential function (which changes the upper half plane to a strip) we can map strips to the unit disc.

Next time we will quickly study linear fractional (or Möbius) transformations, with the assistance of stereographic projection.

1. A little more Argument; locating zeros with Rouché's Theorem
2. The conformal mapping property of analytic functions (why is this interesting?)
3. Linear fractional transformations and the Riemann sphere.

Here are appropriate textbook homework problems for some of the content of the last few meetings of the course.

    3.1:10,11,12,14,18,20; 3.3:4a),b),d);5a),b),c);7a),b),d).

Boundary data tells about roots: the real case
The Intermediate Value Theorem gives some information about what happens inside an interval to a continuous function when certain boundary information is known. Here is how to think of it:
Suppose a< b are real numbers and that f is continuous and real-valued on [a,b]. Then if f(a)f(b)<0, f(x)=0 has at least one real root on (a,b).
Of course f can have lots of roots, and even if f(a)f(b)≥0, f can still have lots of roots. But the idea is there in a primitive form. The Argument Principle gives you much more information.

The Argument Principle
If f is not identically zero with some zeros and poles, and if C is a simple closed curve which avoids the poles and zeros of f, then (1/{2Π i})∫_C[f´(z)/f(z)] dz is

• the net number of poles and zeros (counted with multiplicity) contained inside C.
• the total change in the argument of f(z) as z travels along C.
• how many times f(s(t)) (if s(t) is the parameterization of C) winds around 0 as t goes from START to END.

I mentioned that locally the number of roots counted with multiplicity doesn't change when you consider complex behavior. Restricted to the reals, things can get complicated: consdier x², x²+(small positive), and x²+(small negative). If you only consider real roots and number of points where the func tion is 0, then these "nearby" functions have 1 and 0 and 2 roots. When thinking complex, however, the number of roots counted with multiplicity stays stable at 2 in all of the cases.

An example from the textbook
I will discuss Example 3 from section 3.1:

Let λ>1. Then z+e^–z=λ has exactly one solution in the right half-plane where Re(z)>0.

Mr. Cantave remarked correctly that this equation could be "solved" with the Lambert W function, yet another non-classical special function. For information about it, please see here or here.

The Argument Principle is usually applied somewhat indirectly, although I have seen applications where people actually worry about trips around 0. The usual intermediate result is called Rouche's Theorem.

Rouché's Theorem
Suppose U is a connected open set in C, f and g are holomorphic on U, and C is a simple closed curve whose interior is contained in U. Also suppose that if z is on C, |f(z)|>|g(z)|. Then the number of zeroes of f(z) inside s equals the number of zeroes of f(z)+g(z) inside s.

Heuristic discussion
A supportive discussion via dog-walking will be offered initially.

Proof We know that the number of zeros of f(z) inside C is 1/(2Pi; i)∫_C[f´(z)/f(z)]dz. (Note that f is non-zero on s since |f(z)| is strictly greater than the non-negative number |g(z)| for all points z on C.) We now consider 1/(2Pi; i)∫_C[{f´(z)+tg´(z)}/{f(z)+tg(z)}]dz where t goes from 0 to 1. We call this second integral F(t), and note that it is a continuous function of t (t in [0,1]) that is defined everywhere, since |f(z)|>t|g(z)| on C, so |f(z)+tg(z)|>|f(z)|–t|g(z)|>0, and because the function (t,z)→[{f´(z)+tg´(z)}/{f(z)+tg(z)}] is jointly continuous in z and t. So its integral is continuous.
However, we know that F(t) is always an integer from the Argument Principle, and so it must be the same integer for all t since it is a continuous integer-valued function. Hence, F(1)= F(0), and so the number of zeroes of f+g inside s is the same as the number of zeroes of f inside s.

The classic application of Rouché's Theorem is apparently in solving qualifying exam problems. What unbecoming cynicism! Here is a standard such problem: If |a|>e, show that e^z=azⁿ has n zeroes in |z|<1.

To solve this, we use Rouché's Theorem by taking C to be the unit circle, f to be equal to –azⁿ and g to be e^z. When |z|=1, |f(z)|=|a|, and |g(z)|=e^|Re(z)|1<|a|. So |f(z)|>|g(z)| on |z|=1, and since f(z) has n zeros in |z|<1, f+g must have n zeros also.

Note that here we have used the fact that min|f(z)|>max|g(z)|, where the inf and sup are taken over all z in C. This certainly implies that |f(z)|>|g(z)|. However, the converse is not true, and it is enough to show that the inequality |f(z)|>|g(z)| holds pointwise for z in C for us to apply Rouché's Theorem. I have unfortunately known this "pointwise" estimate to be useful in specific circumstances.

A really terrific UC Berkeley qualifying exam problem
How many roots does the polynomial P(z) = z⁵+z³+5z²+2 have inside the annulus 1<|z|<2?

Take C to be |z|=1, f(z)=5z², and g(z)=z⁵+z³+2. If |z|=1, then |f(z)|=5, and |g(z)|≤|z|⁵+|z|³+2≤4. Since 4<5 we can apply Rouché: the number of zeros of P=f+g has inside this C is 2.

Now, take C to be |z|=2, f(z)=z⁵, and g(z)=z³+5z²+2. Then |f(z)|=32, whereas |g(z)|<=|z|³+5|z|²+2=30<32. So P has 5 roots inside C, and there must be three roots in the annulus.

Note that if |z|≥2, then P(z) never vanishes, since then |z|⁵>|z|³+5|z|²+2.

Linear algebra and angle preservation
Now I hope to do a linear algebra exercise. I will find all linear maps from R² to R² which preserve angles. Examples of such mappings are rotations, dilations (just multiplying lengths), and reflections (such as complex conjugation -- reflection across the real axis).
I will do this with an informal discussion -- a sequence of observations. I will suppose that the linear map has matrix M given by

( a b ) 
( c d )

1. What if det(M)=0?
2. More to come ...

1. Counting zeros and poles with the Argument Principle
2. Locating zeros with Rouché's Theorem
3. The conformal mapping property of analytic functions (why is this interesting?)
4. Linear fractional transformations and the Riemann sphere.

Antiderivatives Here's the problem: suppose U is an open connected set and f(z) is analytic in U. Must there always be F(z) analytic in U with F´=f?

No Suppose U is the "punctured plane", with 0<|z|<∞, and f(z)=1/z. There's can't be F as desired, because if there were, then integrals of f(z) dz over curves C in the punctured disc would be evaluated by antidifferentiation (vai F(end of C)–F(start of C)) and so if C is a all the integrals would be 0. But ∫_|z|=1(1/z) dz=2Πi≠0. There is really only one example in this subject.

Yes It turns out if the domain has no holes ("simply connected") then there is always an antiderivative. The "no holes" property has a complicated name ("simply connected") which I don't want to stress in this course. Instead, let me verify the claim for a disc of radius R>0 centered at p. I did this earlier with another proof when I was discussing Morera's Theorem. So suppose f(z) has a power series expansion, ∑_n=0^∞a_n(z–p)ⁿ, valid for |z–p|<R. Then the series ∑_n=0^∞(a_n/{n+1})(z–p)ⁿ⁺¹ has the same radius of convergence and its derivative is f(z). The sum of that series is F(z).

Logarithms
It turns out that the existence of logarithms of functions is linked to the idea of antiderivatives, as we'll see in 14 seconds. Well, suppose f(z) is analytic in a connected open set U. When does f(z) have a logarithm, let me call it L_f(z)? What's a logarithm? My definition is that L_f is a logarithm if e^L_f(z)=f(z) for all z in U. Does every analytic function have a logarithm?

NO!!!! (but for a silly reason). Just consider the entire function z. I bet there is no entire function g(z) with e^g(z)=z. That's because it would be difficult to imagine a value of g(z) which would make the equation true when z=0, since exp is never 0.

A more subtle NO Consider z in the punctured plane. This is never 0. Suppose we can find L(z) with e^L(z)=z in C\{0}. Then differentiate. We get e^L(z)L´(z)=1, so (since e^L(z) is supplosed to be z!) L´(z)=1/z in C\{0} which we have just seen is impossible.
Notice that I am not asking if any "branch" of log which you happen to know is a logarithm. I am asking a much more difficult question, which is does any one know any function which could be a candidate for "log"(z). The answer is NO so we really didn't lie about logs in any serious way!

Yes A non-zero analytic function in a disc has a logarithm. Why is this true? Suppose f(z) is such a function. If we want e^L(z)=f(z), then L´(z) must be f´(z)/f(z). I can use this observation to try to find a logarithm. Why don't we just antidifferentiate f´(z)/f(z)?

O.k.: suppose g(z) is a function whose derivative is f´(z)/f(z): we know we can find such a g(z) by our earlier observations. What can I then say about e^g(z) in comparison with f(z)? I'll look at e^g(z)/f(z) (this is a quotient of non-zero functions and is therefore not zero). Differentiate this. The quotient rule gives e^g(z)g´(z)f(z)–f´(z)e^g(z) "on top". But g´(z) is supposed to be f´(z)/f(z). Then the top is e^g(z)({f´(z)/f(z)}f(z)–f´(z)). This is 0. So the derivative is 0 and the function (in a connected open set) is constant: e^g(z)/f(z)=Const. The "Const" is non-zero, so I know it is in the range of exp. Find Q so that e^Q=Const. Then the function g(z)–Q is a logarithm of f(z). (Isn't this neat? This is very very very tricky reasoning to me.) By the way, there are lots of logs -- we can add or subtract any multiples of 2Π i.

Now let me seemingly change topics, but things will link up in just a bit.

How to look at rational functions
Suppose R is a rational function. Then we can write R(z)=P(z)/Q(z), where P and Q are polynomials. Since we know that C is algebraically closed, we can (in theory, at least!) completely factor P and Q. We can strike out common factors. We would have a "description" of R which would be multiplicative and unique: R(z) is a product of a constant multiplied by (z–z_j)^N_j, where the N_j's are integers (some positive and some negative).

Example?
Here is a rational function:

  5        4      3      4        3      2        2
 z  – 2 I z  + 2 z  + 3 z  – 6 I z  – 2 z  – 6 I z  – 3 z – 2 I z – 1
---------------------------------------------------------------------- 
                     4      3       2
                    z  + 8 z  + 18 z  – 27

 [(z+1)3(z–i)2]
----------------
 [(z+3)3(z–1)1]

The picture I would like you to see is to the right. It has all (or almost all the information about the rational function, because two rational functions which have the same zeros and poles can differ only by a non-zero multiplicative constant!) of the information about the rational function, but displayed in a rather pleasing and possibly informative geometric manner. The poles of the function are "at" the points which have negative integer weights. The zeros of the function are "at" the points which have positive integer weights. This assemblage (?) is called the divisor of the function (I won't use this word again in this class).

Remember that near a pole or a zero of function we can get a neat description: f(z)=(z–p)^Nh(z) where if N is positive, we have a zero of order N and if N is negative we have a pole of order –N. The function h(z) is not zero near p. Notice that f´(z) is N(z–p)^N–1h(z)+(z–p)^Nh´(z). So f´(z)/f(z)=[N(z–p)^N–1h(z)+(z–p)^Nh´(z)]/[(z–p)^Nh(z)] and this simplifies to N/(z–p)+h´(z)/h(z). This is a function with an isolated singularity at p, whose residue is N. So the Residue Theorem will count the N's.

Then applying the "construction" above to the rational function I gave, the residues of the derivative divided by the function -- we seem to get a geometric picture of 4 points, 1 and –i and –3 and –1, together with integer weights at each of the points. The integers represent zeros with their order or multiplicity and poles with their order or multiplicity.

So suppose we have function f and a simple closed curve C, oriented counterclockwise as usual, and suppose that the curve does not go through any of the poles and zeros of C. Let's have s(t) defined on the interval [a,b] be a parameterization for C. What is (1/{2Π i})∫_C[f´(z)/f(z)]dz? One interpretation has something to do with logs.

We cover C by a finite collection of open discs, disjoint from the poles and zeros of f. In each disc, we get L_j(z), which is a version of log(f(z)). We get the antiderivative as usual by taking values of L_j (j going from 0 to M, say) at two points in each disc. So the value of the integral is ∑_j=1^ML_j(s(t_j))–L_j(s(t_j–1)) where s(t) is a parameterization of C. Notice, please that the sum "telescopes" in the sense that s(t_M)=s(b)=s(a)=s(t₀). Notice further that L_j(t_j)–L_j–1(t_j) must be 2Π i multiplied by an integer since we are subtracting two "determinations" of logarithm. So all this tells me that the integral will be 2Π i multiplied by an integer. The curve f(s(t)) for t between t_j and t_j+1 can't go through 0 (since f is away from its zeros on C) and the imaginary part of the difference between the values of L_j represents the change in the argument of f(s(t)) in the piece of C inside the j^th disc. The total of all of these changes in argument is 2Π i multiplied by an integer, and this integer, if you ponder it carefully, is how many times the curve of the curve f(s(t)) winds around around 0. That is the value of the sum! Therefore the integral also evaluates the change in the argument of f around the curve C.

The Argument Principle
If f is not identically zero with some zeros and poles, and if C is a simple closed curve which avoids the poles and zeros of f, then (1/{2Π i})∫_C[f´(z)/f(z)] dz is the net number of poles and zeros (counted with multiplicity) contained inside C. It is also how many times f(s(t))(if s(t) is the parameterization of C) winds around 0 as t goes from START to END. Why? Because if you look at the integral, and make the "substitution" u=f(z), then the integrand becomes {1/u}du, and the curve integrated over is f(s(t)), the image of s under f. Therefore the integral measures the change in the argument of f as we "travel" around s, Δ_C(arg f).

An example (!!)
O.k., this is a bit risky, but let's try to see an example at work. We've got good technology, and things should be visible. So let's use the function f(z) discussed earlier:

 [(z+1)3(z–i)2] 
----------------
 [(z+3)3(z–1)1]

Fact An analytic function f(z) has a zero of order N at p (where N is a positive integer) exactly when f(z)=(z–p)ⁿh(z), where h(z) analytic near p and h(p)≠0,
or
f(p)=0, f´(p)=0, f´´(p)=0, ..., f^(N–1)(p)=0, and f^(N)(p)≠0.

Since we know that the function has a Taylor series expansion centered at p, and since we know that the j^th coefficient is f^(j)(p)/j!, I really don't think this fact is difficult to check.

Since h(z) is analytic, it must be continuous. Since h(p)≠0, there is some δ>0 so that h(z)≠0 for 0<|z–p|<δ (take ε=(1/2)|h(p)| and make z so that if |z–p|<&delta, then |h(z)–h(p)|<ε). So inside a disc of radius δ the only z where f(z)=0 is z=p. The zeros are isolated. A consequence of this harmless observation is the following startling fact about the "geometry" of all of the zeros of f(z).

No crowding!
Suppose f(z) is analytic in an open, connected set U. Suppose f(z_n)=0 and lim_n→∞z_n=p in U. Then f must be the zero function in all of U.

This is an amazing and almost unbelievable result, so let me prove it before discusing examples and consequences.

Since f is analytic, it must be continuous, so f(p)=f(lim_n→∞z_n)=lim_n→∞f(z_n)=0. So p is a zero of f(z), and therefore if f(z) is not always 0, we have the description we mentioned previously: f(z)=(z–p)^Nh(z) for some positive integer N and for some function h(z) analytic near p with h(p)≠0. But f(z_n)=(z_n–p)^Nh(z_n) so since the first factor is not 0, h(z_n) must be 0. This is a contradiction, since h(p) is the limit of the h(z_n)'s which are all 0. Therefore f must be the identically 0 function near p.

But "identically 0" spreads around the connected open set U. I gave this argument at least once earlier in the course -- let me run rapidly through it again. Let's draw a line inside U between p and q. Since f s 0 near p, the power series expansion for f is 0 inside a disc centered at p. f and all of its derivatives vanish in that disc. Now select another point in the disc, moving from p towards q. Since f and all of its derivatives vanish at that point, I know that the coefficients of the power series coefficients for f at that point are all 0. So in a disc centered at that point, f must be identically 0. I can continue along a chain of discs from p to q, and then f must be 0 in a disc which includes q.

But "connected" means I can go from p to any other point in U using a finite number of such line segments, and therefore "being zero" spreads from a disc around p to all of U.

Repeating ...
The conclusion of the theorem: there is no such crowding of zeros unless f(z) is always 0!

Example
Consider the analytic function sin(1/(z–1)) in the unit disc, |z|<1, certainly open and connected. Then f(1–{1/[2nΠ])=0, and if z_n=1–{1/[2nΠ], certainly this sequence of zeros piles up at 1. The picture to the right is an attempt to show the unit disc together with the location of just a few of the infinitely many zeros of the is function (attempts to show all of the infinitely many zeros in some clear fashion seem to be difficult).
But f(z) is not always 0, so we seem to have a counterexample to the theorem which was just proved. What's happening? ... There is not a problem. The limit point of the zeros, 1, is not an element of the domain, and therefore the hypotheses of the theorem are not satisfied. The example is irrelevant.

A slight restatement of the result just proved is frequently called the Identity Theorem: two functions, f(z) and g(z), both analytic on a connected open set U, which agree on a subset of U with an accumulation point, must agree on all of U. So tiny pieces of information are ENOUGH!!! We could conceivably prove using, say, mathematical induction, that f(z_n)=g(z_n) on some seuqence with a limit point in U, and then, even though we don't realize it, we've shown that f and g are the same on every point of U, no matter how "far away". Amazing! Well, let me show you some examples.

I know a function which is its own derivative and whose value at 0 is 1. That's e^x. So any analytic function f(z) which equals e^x on, say, the real line, must be e^z. And any analytic function which equals e^z must be its own derivative. Sigh. Differential equations can be "complexified" -- just check them on R and, if true there, they must be true on any connected open set U which contains even a short segment of R. Or even just a sequence with a limit point!

Suppose g(z) and h(z) are two entire functions which are equal to sin(x) and cos(x), respectively, on R. Then for all z, I know that g(z)²+h(z)²=1. Indeed. So I know that (good grief!) [sin(x)cosh(y)+i cos(x)sinh(y)]²+[cos(x)cosh(y)-i sin(x)sinh(y)]²=1 for all x and y without any further checking. Both squared things are analytic functions and they are sin(x) and cos(x) when y=0 so the equation is correct when y=0, a subset of C which has lots of limit points.

So there is a persistence of "identities" as well for algebraic equations.

Here is a more subtle problem. Consider the function f(z) defined by the formula f(z)=e^z/{(z+3)(z⁴+1)}. f(z) is a function which is analytic with isolated singularities at 3 and at the fourth roots of –1 (this are, I think, (±1±i)/sqrt(2). Since in each case we can write an appropriate factorization in terms of (z–CENTER)^–1 times h(z) with h(CENTER)≠0, we know that these singularities all happen to be "simple poles", or poles of order 1.

General theory states that this function has a power series centered at 2 with a radius of convergence equal at least to the distance from 2 to the nearest singularity (or point in C which is not in the connected open set in which f(z) is analytic. So I guess nearest to 2 is (1±i)/sqrt(2). The distance of either of these points to 2 is the same, and is sqrt(sqrt(2)²+(2–1/sqrt{2})²) which is sqrt({11/2}–2sqrt(2))≈1.63449. So that's an underestimate of the radius of convergence. Can we get more precise information?

In fact, the radius of convergence is actually equal to Q=sqrt({11/2}–2sqrt(2))≈1.63449. Why is this? Well, suppose f(z) is f(z)=e^z/{(z–3)(z⁴+1)}. Then let g(z) be the sum of the power series of f(z) centered at z–2. We know that g(z) converges in at least a disc centered at 2 with radius sqrt({11/2}–2sqrt(2)). Suppose it converged with a larger radius. Let T=(1+i)/sqrt(2), one of the closest (to 2) singularities of f(z). Well, then since g(z)=f(z) inside the disc of radius Q centered at 2, I know that lim_z→Tg(z)=lim_z→Tf(z) (if we go through z's for the limit which are inside the disc of radius R centered at 2. But sinde f(z) has a pole at T, I know that lim_z→Tf(z) does not get close to a complex number (which would be the candidate for g(T) since the f limit blows up (|f(z)|→∞). So the g limit can't exist, and the power sries sum can't converge in a larger disc. We need the identity theorem to be sure that the sum of the Taylor series for f, which is g(z), converges to f(z) inside the entire disc of convergence, not just "close" to the center which Taylor's Theorem etc. might guarantee.

Notice that although we now have very precise information about the radius of convergence, direct computation of the Taylor coefficients is not feasible. Less politely, it is a mess! For example, a pal of mine tells me that the coefficient of (z–2)¹⁵ is ({48809 35384 43023 38924 81049 95899 53214 88149}/{10564 07115 86559 56210 25706 55775 00000 00000 00000})·e². As Hamming declared, "The purpose of computing is insight, not numbers." Here is a computation which gives no insight. It is the theoretical discussion which gives some insite.

By the way, once the radius of convergence is known, then statements about the asymptotic growth rate of power series coefficients can be made. This is used in real applications.

This result about radius of convergence is not true in "real calculus". So rather bizarre examples are possible such as the "kinky" (?) example we considered earier in the course (e–1/x if x≥0, 0 otherwise). That function is equal to its Taylor series centered at –1 in the interval (–∞,0), but the equality does not extend to larger intervals. This can't occur in complex analysis.

The Gamma Function
The Gamma function is one of the "well known" (!) special functions in mathematics. It is supposed to be the most natural interpolation and extension of the usual factorial defined for positive integer values. Here is the official definition:
    Γ(x)=∫_t=0^∞t^x–1e^–tdt.
Then Γ(n)=(n–1)! when n is a positive integer. This is verified with integration by parts. I've done the computations in several calc 2 classes -- they are not difficult. But here we're supposed to be doing complex calculus. Certainly if Re(z)>1, the improper integral ∫_t=0^∞t^z–1e^–td converges. That's because
    |t^z–1|=|e^(z–1)ln(t)|=e^{Re(z–1)ln(t)}=e^{[Re(z)–1]ln(t)}=t^(Re(z)–1)
and then e^–t decreases rapidly enough to control the growth at ∞ of the t power. We need Re(z)>0 so that the integral of the t power is finite at 0 (even if improper). The trick we've used with Morera's Theorem shows that Γ(z) defined by this formula is analytic in Re(z)>1. But amazingly Γ(z) has an alternative formula for other z's, and then by analytic continuation the function can be defined on larger domains. Since the overlap has accumulation points, I know that everything works well by the Identity ideas. What's the other formula? Here:

Since Γ(x)=∫_t=0^∞t^x–1e^–tdt we integrate by parts: u=e^–t and dv=t^x–1dt. Then du=–e^–tdt and v=t^x/x. Take limits etc. (these are improper integrals!) so then Γ(x)=(1/x)Γ(x+1). The right-hand side is defined for x>–1 except at 0. So...
Γ(z) now is extended to be analytic in Re(z)>–1 except for a simple pole at 0.

The Residue Theorem can be used to verify facts about the convergence and sum of certain infinite series. For example, it can be used to show that ζ(2)=∑_N=1^∞1/n²=Π²/6 using the function f(z)=cot(Πz)/z² (not a joke!). The computation is proof #8 in this reference which has fourteen (that's 14) proofs of the equality. This proof is called a "textbook proof, found in many books on complex analysis.". There's just not enough time!

Now I want to move on to a related topic. But first, some background:

What does a pole look like?
Suppose f is analytic in 0<|z–p|<R. Then f has a Laurent expansion: f(z)=∑_n=–∞^&infina_n(z–p)ⁿ. Let's suppose that f does not have an essential singularity at p. So it has either a pole or a removable singularity. If the first alternative is true, then f(z) is locally (z–p)^Nh(z) where h(z) is analytic in |z–p|<R with h(p)≠0. We get this by "factoring out" (z–p)^N, so h(z)=∑_n=N^&infina_n(z–p)^n–N with h(p)=a_N≠0. This is a pole of order –N (remember N is negative here!).

This silly result actually imposes rather severe restrictions about the behavior of f near an isolated singularity. Look at these cliams:

Exam question #1
Suppose f has an isolated singularity at 0, and we know the following two facts:
    i) If n is a positive integer, then f(1/n)=n²(4–33i).
    ii) If n is a positive integer, the f(i/n)=n⁷(5+2i).

Then I claim f has an essential singularity at 0. Certainly the singularity can't be removable since lim_n→∞f(1/n) is not a complex number. Can the singularity be a pole? Well, if it were a pole, then there would be a negative integer N and a non-zero constant W so that f(1/n) would appear to be Wn^–N. O.k., i) identifies candidates: N=–2 and W=4–33i. But ii) yields N=–4 and W=5+2i. The N and W should be the same so 0 is not a pole. It must be essential.

Exam question #2
Suppose f has an isolated singularity at 0 and I know that when n is a positive integer, f(1/n)=nⁿ. Then I claim f has an essential singularity at 0.

Here there is no fixed N whihc would give nⁿ=(1/n)^N. So (same reasoning: neither removable nor a pole, so 0 must be essential.

Now let's look at removable singularities
Notice that if f has a removable singularity at p, then we can assume that f(p) is defined and f is analytic in |z|≤R. We can do something resembling the factorization for poles, only here the N is positive. If f is not always 0, there is a smallest N so that a_N≠0. Then write f(z)=∑_n=–∞^&infina_n(z–p)ⁿ=∑_n=0^&infina_n(z–p)ⁿ=∑_n=N^&infina_n(z–p)ⁿ=(z–p)^Nh(z), where h(z)=∑_n=0^&infina_n+N(z–p)ⁿ and h(p)=a_N≠0.

Exam question #1 (revised!)
Suppose f has an isolated singularity at 0, and we know the following two facts:
    i) If n is a positive integer, then f(1/n)=(1/n)²(4–33i).
    ii) If n is a positive integer, the f(i/n)=(1/n)⁷(5+2i).

Then I claim f has an essential singularity at 0. Certainly the singularity can't be a pole since f(1/n) doesn't →∞ as n→∞. If f had a removable singularity, then the value of f at 0 would be 0 since the limit of {f(1/n)} is 0. But the rates are different. i) suggests that the zero has order 2, adn ii) suggests that the zero has order 7. These descriptions shold be consistent, and they are not.

Exam question #2 (revised)
Suppose f has an isolated singularity at 0 and I know that when n is a positive integer, f(1/n)=1/(nⁿ). Then I claim f has an essential singularity at 0.

Here there is no fixed N which would give (1/n)ⁿ=(1/n)^N. So use the same reasoning: neither removable nor a pole, so 0 must be essential.

If N>0, then N is called the order of the zero of f at p. What can we say about this? A polynomial like (z–i)⁴z(z+7)⁵ has zeros at i, 0, and –7 respectively, of order 4, 1, and 5. The h's with non-zero values near each zero are easy to write (just the other factors of the polynomial).

Can an analytic function have infinitely many zeros? Well, yes (consider sin(z), for example!). I claim that sin(z) has zeros of order 1 at each nΠ (n is an integer). Why? Well, if we write sin(z)=(z–nΠ)h(z), we can differentiate, and see that cos(z)=h(z)+(z–nΠ)h´(z). At nΠ the left side of the equation is not 0, and the right side of the equation is h(nΠ), so h(nΠ) is not 0. In fact, this immediately allows us to realize that a function like [sin(z)]¹⁷ has zeros of order 17 because since sin(z)=(z–nΠ)h(z) where h(z) is analytic and h(nΠ)≠0, we have [sin(z)]¹⁷=(z–nΠ)¹⁷k(z) where k(z) is analytic and k(nΠ)≠0 (indeed, k(nΠ)=[h(nΠ)]17≠0).

The computation I did for sine above is actually a bit more general. Here is a result which is sometimes useful.

Fact An analytic function f(z) has a zero of order n at p (where n is a positive integer) exactly when f(z)=(z–p)ⁿh(z(), h(z) analytic near p and h(p)≠0, or f(p)=0, f´(p)=0, f´´(p)=0, ..., f^(n–1)(p)=0, and f⁽ⁿ⁾(p)≠0.

Analytic functions can even have zeros of unbounded order (construction of this is not so simple). Although such a function can have infinitely many zeros, there is a slightly subtle restriction on the geometry of the zeros.

Residue examples

The Fourier transform of 1/(1+x2) ∫–∞∞[cos(mx)/(1+x2)]dx Maple says this should be –Πsinh(|m|)+Πcosh(|m|). Here the "surprise" is that we do not take f(z) to be cos(mz)/(1+z2) because that would not have growth which would be easy to estimate on the upper semicircle. Instead, for m>0, take f(z)=eimz/(1+z2). This f(z) has a simple pole at i, so f(z)=h(z)/(1-i). Then res(f;i)=h(i).
∫–∞∞[ln(x)/(1+x2)2]dx Here we needed an indented contour to take care of problems at 0. The integrand had to be chosen with some care in order to get a good definition of log. We took f(z)=Log(z)/(1+z2)2. f(z) has a double pole at i, and if f(z)=h(z)/(z-i)2, then res(f;i)=h´(i). This should be –Π/4 (with 1+x2 downstairs the integral is 0 -- use a cute change of variables: x→1/x to see this "clearly").
∫0∞dx/(1+x4) (Two simple poles. With luck skill, the result is real and even positive.) This should be Πsqrt(2)/4. My hurried computation did not quite get this, I think. The residue computation was gotten (if the simple pole is at, say, r) by looking at limz→r(z-r)/(1+z4) and using L'Hop.
∫02Πdθ/(5+3sin(θ)). Clearly (NO!) change the integral into an integral around the unit circle. We already did this integral in lecture #??. You can inspect what we did and find the Residue Theorem "lurking" in the computation.
∫0∞dx/(1+x3) Many coincidences. Choosing the correct contour works, but it is emphatically not obvious! This should be 2Πsqrt(3)/9.

There are whole books about the use of the Residue Theorem in computing integrals (proper and improper) and in computing infinite series.

Mr. Cantave remarked after class that it seems all the examples I showed could be done in various ways without the Residue Theorem itself. I had looked at poles, where the results could maybe have been gotten with various manipulations involving the Cauchy Integral Formula. That is certainly correct as far as the examples we've seen. But there are other more complicated situations where the Residue Theorem itself is a more appropriate tool. I don't have time to show you those.

Example 2
Let's consider f(z)=z/sin(z). I know that sin(z)=sin(x+i y)=sin(x)cos(i y)+cos(x)sin(i y)=sin(x)cosh(y)+i cos(x)sinh(y). Note that since cosh(y) is never 0, this is 0 only when sin(x)=0. Note that to get the imaginary part=0, we need y=0 since cos(x) will never be 0 when sin(x)=0. So the zeros of sin(z) are when z=(integer)Π.

Suppose n is an integer. What does the power series for sine look like at nΠ? Since sin(nΠ)=0 and cos(nΠ)≠0 (yeah, yeah, I'm too busy to notice whether n is even or odd -- I just care if things aren't 0 right now!). So sin(z)=0+A(z–nΠ)+0(z–nΠ)²+B(z–nΠ)³+H.O.T.. There are now two cases:

• n=0. Here f(z)=z/[0+Az+0z²+Bz³+H.O.T.]=1/[A+0z+Bz²+H.O.T.] and what is on the bottom is a convergent power series because I got it by factoring out a z from a convergent power series. Therefore lim_z→0f(z)=1/A≠0 so f(z) has a removable singularity at 0.
• n≠0. Now f(z)=z/[0+A(z–nΠ)+0(z–nΠ)²+B(z–nΠ)³+H.O.T.]. I notice that as z→nΠ, the top→nΠ≠0, while the bottom→0. We can write f(z)=(z–nΠ)^–1h(z), where h(z)=z/[A+0(z–nΠ)¹+B(z–nΠ)²+H.O.T.] is a function analytic near nΠ and h(nΠ) is (nΠ/A)≠0. So this f(z) has a pole of order 1 (frequently called a simple pole [I will make an appropriate joke about this at some time!]) at nΠ.

Example 3
Take f(z)=e^1/z. Since I know its Laurent series (take ∑_n=0^∞zⁿ/n! and stuff in 1/z for z to get ∑_n=0^∞{1/n!}z^–n, which certainly has infinitely many non-negative coefficients. But also I know that I can solve e^1/z=b (for b≠0) whenever 1/z=log(b). This means 1/z=ln(|b|)+i Arg(b)+2Πn i, so that z=1/[log(|b|)+i Arg(b)+2Πn i]. Since the sequence log(|b|)+i Arg(b)+2Πn i→∞ as n→∞, I know that the sequence of reciprocal entries→0 as n→0.

The pictures illustrating this are to the left and to the right.
On the left Those are supposed to be some representatives of the set log(b). There's a specified Log(b) with some colors, and the other elements are translations up and down by 2Π i of Log(b). There are infinitely many of them, of course.
On the right These complex numbers are reciprocals (1/z) of the numbers in the first picture. It happens that a vertical line becomes (under the transformation z→1/z) a circle going through 0 (I'll show you soon a rapid way to verify that instead of using computation). As the dots in the left picture sail up and down by multiples of 2Π i, the images in the right are a sequence which →0 in two ways.

This illustrates a strong version of the qualitative consequence of Casorati-Weierstrass. The values of e^1/z are not only are close to b's, but they are equal to b's (with one exception, b=0).

This example is not an accident. There is a much more "powerful" result called The Great Picard Theorem which asserts that if f(z) has an essential singularity at 0, then on any open neighborhood of 0, f(z) takes all possible values, with at most one possible exception. For e^1/z, the exceptional value is 0. The Big Picard Theorem is difficult to prove.

Discussion question
Where does f(z)=e^1/[sin(1/z)] have isolated singularities, and what kind of singularities are they? Well, the function is not defined where sin(1/z)=0 or where 1/z is not defined, and that's where z=1/[nΠ] and z=0. So candidates are those numbers. But (trick question!) z=0 is not an isolated singularity! It is not isolated, since it is a limit point of other places where f(z) is not defined. All of the 1/[nΠ] points are isolated singularities, and they are essential.

Baby Residue Theorem
Suppose f is analytic in 0<|z–p|<R and C is a simple closed curve in that disc with p inside C. Then ∫_Cf(z) dz=2Π i a_–1, where a_–1 is the coefficient of (z–p)^–1 in the Laurent series for f in 0<|z–p|<R.

First write out the Laurent series for f, and then interchange integration and summation (this is valid because we have nice error estimates for the series). Then notice that all the other pieces of the Laurent series, C(z–p)^N for N≠–1 are derivatives of analytic functions, C(z–p)^N+1/(N+1), and we know that the integral of g´ around a closed curve is 0. Then we deform C into a small circle around p and get just our "usual suspect", a_–1∫1/(z–p) dz, giving us the desired result 2Π i a_–1. We could also think, instead of a deformation, that there is sort of a crosscut going back and forth between the small circle and C. Then "inside" the contour gotten by going around C, down the crosscut to the circle, backwards around the circle, and up again using the crosscut, we have a region in which f is analytic, so (Cauchy's Theorem!) the integral is 0. But the crosscut integrals (in green in the picture) cancel each other, and then the integral around C minus the integral around the circle (backwards) have total value 0. So the two integrals are equal.

This is a prototype for one of the most interesting results in mathematics, the Residue Theorem. Your text calls this a_–1 the Residue of f at p, and written Res(f;p). Please: other texts and references have different notations.

More grownup Residue Theorem
Suppose C is a simple closed curve in a connected open set U, and f is analytic on and inside C except possibly for a finite number of isolated singularities inside C. Then ∫_Cf(z) dz=2Π i ∑_{p isol. sing. inside C}Res(f,p).

This result has been tremendously influential in mathematics and its applications. One reason is that it compares global and local information. The integral is some sort of "global" information about f on C. The Residue Theorem says that this equals a sum (a finite sum) of "local" data about f, these residues.

Proof There are many very elaborations of this result, Here in Math 403, I see the verification of the result is a consequence of a picture which is displayed to the right. There is C, and there are displayed the isolated singularities of f inside C. Look at the picture, consider the crosscuts and realize that we're done: each of the circles contributes the residue at the center of the circle, and the crosscut integrals all cancel out. We're done!
Secret strategy We're using different Laurent series representations around and inside each circle!

Workshop problem
Now the instructor will distribute the workshop problem and students will work in small groups to complete the problem.

It was a wonderful occasion. Many people were able to do almost all of the workshop. Let me discuss the solutions in some detail.

1. f(z)=1/(z²+z+1).
2. Use the Quadratic formula. A=-1/2+[sqrt(3)/2]i and B=-1/2-[sqrt(3)/2]i. Since the bottom of f(z) becomes 0 at A, my bet is that f(z) has a pole at A.
3. f(z)=1/[(z-A)(z-B)].
4. Near A I think this way about f(z): it is 1/(z-a)·h(z), where h(z)=1/(z-B) is certainly analytic near B and h(A)≠0, so f(z) has a simple pole or a pole of order 1 at A. Since I know h(z) is analytic near A, I know h(z)=h(A)+h´(A)(z-A)+H.O.T.. Then f(z)=[1/(z-A)]{h(A)+h´(A)(z-A)+H.O.T.} and I only care about the coefficient of 1/(z-A). The (z-A) powers of order 1 and more don't matter, so Res(f,A)=h(A)=1/(A-B)=1/(sqrt(3) i).
5. The Residue Theorem applies and the value is 2Πi[1/(sqrt(3) i)]=2Π/sqrt(3).
6. |f(z)≥|z²|-(|z+1|)≥|z|²-|z|-1=R²-R-1.
7. The length of S_R is ΠR. ML applied to |∫_SRf(z) dz|≤(ΠR)/(R²-R-1).
8. The limit is 0.
9. The limit is 2Π/sqrt(3).
10. ∫_-∞^∞[1/(x²+x+1)] dx=2Π/sqrt(3).

Now we consider 3z/(z–4)². We again rewrite in terms of z–i:
3z/(z–4)²={3(z–i)+3i}/{(z–i)–(4–i)}²=[3/{(z–i)–(4–i)}]+[{3i+3(4–i)}/{(z–i)–(4–i)}²]. Once again, there are two cases, |z–i|<|4–i|=sqrt(17) and |z–i|>|4–i|=sqrt(17).
For |z–i|<|4–i|=sqrt(17), we analyze 1/{(z–i)–(4–i)} first. The result is ∑_n=0^∞–{1/(4–i)ⁿ⁺¹}(z–i)ⁿ. Now I'll give the analysis approach to getting something with 1/{(z–i)–(4–i)}². The derivative of the series tells me:
–1/{(z–i)–(4–i)}²=∑_n=0^∞–{n/(4–i)ⁿ⁺¹}(z–i)^n–1 and so 1/{(z–i)–(4–i)}²=∑_n=0^∞{n/(4–i)ⁿ⁺¹}(z–i)^n–1=∑_n=1^∞{n/(4–i)ⁿ⁺¹}(z–i)^n–1=∑_n=0^∞{(n+1)/(4–i)ⁿ⁺²}(z–i)ⁿ.
Another way, totally correct, is to square the original series and collect terms. This is what's usually done in applications related to combinatorics or perhaps probability, which such series occur in connection with generating functions.

Now, equipped with these results, I can try to write the Laurent series in each annulus, and identify as I do f_in and f_out.

Let's consider the function e^z+(1/z) which is certainly analytic in C\{0}. It is analytic in an open disc centered at z=0 except at the center of the disc. This situation, after a definition next time, will be called, "f has an isolated singularity at 0."

Therefore f(z) can be written as ∑_n=–∞^∞a_nzⁿ. By direct manipulation of the power series for exp, at z and at 1/z, I "found" the (–1)^st coefficient of e^z+(1/z). So a_–1=∑_k=0^∞1/(k!(k+1)!) which Maple identifies as a value of a certain Bessel function. Does this help? Finding every coefficient of a Laurent series explicitly for a random function, even one defined by a classical formula, can be very difficult or even impossible. The (–1)^st coefficient in many, even most, situations turns out to be the only one of serious interest to people. By the way, finding this specific a_–1 is an exercise in one of the 20^th century's most lovely complex variables textbooks, Analytic Functions by Stanislaw Saks and Antoni Zygmund. This book is almost completely different in the order of its subject matter from anything which preceded or followed it.

Laurent series results, specialized for this lecture
Suppose f is analytic in the punctured disc centered at 0 with radius R (that is, the collection of z's with 0<|z|<R. Then

• f(z)=∑_n=–∞^∞a_nzⁿ for all z with 0<|z|<R. This series converges absolutely for all such z, and we can get good convergence estimates on circles inside the annulus.
• a_n=(1/[2Π i])∫_|z|=t(f(s)/sⁿ⁺¹)ds, for any t between 0 and R (actually, for all such t: the integral does not depend on these "eligible" t's!).
• f=f_in+f_out, where f_in is analytic in the whole disc |z|<R and f_out is analytic in C\{0}. (f_in is the sum for $n≥0 and f_out is the sum for n<0).

For me, it is important to note that we "developed" the material in the following table sequentially. We did not present it as all done at once, and obvious by some sort of divine inspiration. It certainly is not that. It was done by human beings, fallibly, with many examples and attempts with varying success.

The entries in the table will be stated for a function having an isolated singularity at p. The supporting discussion will simplify this -- in that, I'll change p to 0 so that I will write less.

We assume that f(z) is analytic in 0<|z–p|<R so f(z)=∑n=–∞∞an(z–p)n there.
Classification of isolated singularities
Name of singularity	Laurent series	Theorem or behavior	Examples
Removable singularity	The an=0 if n<0, so that f=fin.	Riemann Removable Singularity Theorem If f is bounded in 0<|z–p|<R, then there is F analytic in |z–p|<R with F=f if |z–p|>0.	All of these have p=0: z3 (silly, but an example!); [sin(z)]/z, [cos(z)–1]/z2.
Pole	There is N<0 so that aN is not 0, and an=0 when n<N	f(z)=(z–p)Nh(z) for some negative integer N, where h is analytic in a neighborhood of p with h(p) not 0. This is a pole of order N. limz→pf(z)=∞.	(Discussed in more detail below.) Suppose P and Q are in C[z} (polynomials) and that P and Q have no common factor -- remember by the Fund. Thm of Alg., we "can" factor all polynomials. Then P/Q has a pole at every singularity (where Q(z)=0). 1/[ez–1]: this has a pole at 0 if you consider the power series of exp, factor out a z, and then recognize (?) h(z). The behavior is the same at 2Π n i, for all integer n. Take the mth (integer) power of the preceding example. The result certainly has poles of order m at all 2Pi; n i.
Mysterious black hole!!! Essential singularity	There are infinitely many an's with n<0 which are not 0.	Casorati-Weierstrass Theorem For all ε>0, the collection of values f(z) for 0<|z–p|<ε gets arbitrarily close to any w in C.	e1/z and many others.

Riemann's Theorem on Removable Singularities If f is bounded in 0<|z|<R then f can be extended analytically to all of |z|<R.

Proof (There is a different proof in the textbook.) The boundedness condition satisfied by f implies that there is M>0 so that |f(z)|≤M for all z with 0<|z|<R. Then |a_n|=|(1/[2Π i])∫_|z|=t(f(s)/sⁿ⁺¹)ds|. Of course we will estimate the modulus of the integral with ML (what else?). So the modulus is ≤[(2Π t)/(2Π)]M/tⁿ⁺¹=M/tⁿ.
But if n<0, the final expression→0 as t→0. So all of those a_n's are 0, and f=f_in for 0<|z|<R, and f_in is the desired extension.
This is very much like the Liouville's Theorem proof. As are other proofs ... as you will see.

Counterexample to show that this theorem does not work on R
Consider sin(1/x). This is bounded. It is differentiable in R\{0} but it cannot be extended by any sort of continuity at 0. Why doesn't this contradict the result? (Of course, you should consider what happens in "imaginary" directions!)

About poles
The limit means the following statement:
    For all M>0, there is ε>0 so that |f(z)|>M when 0<|z|<ε.

Why does this limit statement hold if we know f(z)=z^Nh(z) with h(z) analytic and h(0)≠0? well, to get |z^Nh(z)|>M we can first consider h(z). It is analytic at 0, hence continuous at 0, so I can get close enough (say |z|<FROG) so that then |h(z)–h(0)|<(1/2)|h(0)| (remember h(0)≠0!). Then |h(z)|>(1/2)|h(0)| so |z^Nh(z)|>(1/2)|h(0)| |z^N|. If I want the underestimate to be >M, well, look:
    (1/2)|h(0)| |z^N|>M ⇔ |z^N|>2M/|h(0)| ⇔ |z|^N>2M/|h(0)|
But N is a negative integer, so making |z|^N large is the same as making |z| small. The inequality above is equivalent to requiring that 0<|z|<(2M/|h(0)|)^–1/N. If I also ask that |z|< FROG, then indeed |h(z)| is guaranteed to be larger than M.

Now how do I get the statement about Laurent series? Since h(z) is analytic in |z|<R, there is a power series equal to h(z). So it looks like this: h(z)=A+Bz+Cz²+... with h(0)=A≠0. We get a series for f(z) by multiplying by z^N. Remember that N is a negative integer. So the series Az^N+bz^N+1+Cz^N+2+... is a Laurent series with a finite number of non-zero negative terms.

If f(z)→∞ as z→0, then 1/f(z) is non-zero for z close enough to 0 (this is a consequence of the definition of f(z)→∞). We look at 1/f(z) and it must →0, so 1/f(z) is bounded near 0. The Riemann Removable Singularity Theorem takes over: we can extend f(z)'s definition to be 0 at 0, and then the extension is analytic. This extended function has a power series. The series can't be identically 0, because then it couldn't be 1/f(z) for z≠0. So it looks like ... well, a series that has a 0 constant term. So: there must be a positive integer M and a≠0 so that 1/f(z) must be equal to az^M+bz^M+1+cz^M+2+... (this is called a zero of order M). Then 1/f(z)=z^M(a+bz+c²+...). What's in the parentheses is a convergent power series with a non-zero constant term. So it must be analytic with sum g(z), and g(0)≠0. Therefore f(z) itself is equal to z^–M(1/g(z)). This is exactly the description we started from!

But please notice that if |f(z)|→∞, it must "go to ∞" in a very precise fashion: essentially proportional (constant of proportionality: h(0)) to an inverse integer power of |z|. This is a very "stiff" requirement.

About essential singularities
The astonishing Casorati-Weierstrass Theorem Suppose that there are infinitely many a_n's with n<0 which are not 0. Then for all ε>0, the collection of values f(z) for 0<|z|<ε gets arbitrarily close to any w in C.

Proof Suppose the conclusion is false. Then there is ε>0, a in C, and b>0 so that for all z in with 0<|z|<ε, |f(z)–a|>=b.
If you can contradict a complicated quantified statement, then you too can prove the Casorati-Weierstrass Theorem!

Consider g(z)=1/(f(z)–a). Then |g(z)|<1/b, so that g is bounded in 0<|z|<ε, and we can apply Riemann's Removable Singularity Theorem. We know that there is G analytic in |z|<ε which extends g. Consider the behavior of f(z)=(1/g(z))+a.
• If G(0) is not 0, then f has a removable singularity at 0, and the Laurent coefficients a_n for n<0 are all 0. This is false.
• If G(0)=0, then (since g is not identically 0) f has a pole of some order at 0, and only finitely many of the a_n's for n<0 are non-zero. This is also false.
This contradiction shows that the original assumption must be false.

We will consider some examples.

What simple examples of entire functions do we know? Even a very short list would include polynomials and the exponential function. What are the values of a typical (non-constant!) polynomial? Well, given w, can we find z so that P(z)=w? Sure -- that follows from FTA. So the range of any non-constant polynomial is all of C. What about exp? That is, for which w's can we cold exp(z)=w? We spent some time on this at the beginning of the course, and there we called roots of this equation values of log(w). Any non-zero w can be "fed into" the log. So the range of exp is all of C except 0. We have had other examples, like sine. Our study of sine, if you look back at it, shows that the range of sine is all of C. Historically people looked at example after example. They kept coming up with the results (for non-constant entire functions!) that the range was either all of C or all of C except one value. This is true, and was a major result of complex analysis about 120 years ago by Picard (see here for some discussion). Although the wikipedia article asserts that "This theorem is not difficult to prove" I do not agree. I've only discussed the proof (or, rather, proofs -- there are more than one) in a second graduate course in complex analysis.

The tools we're going to get in this lecture are used in proving the various theorems of Picard. So let's get to work.

The most important single technique for analyzing holomorphic functions in circularly symmetric domains is the Laurent series, a generalization of power series. A biography of Pierre Alphonse Laurent (1813-1854) declares that "his memoir was not published" (referring to the paper in which the Laurent series was first described). This is not so nice. Laurent series will be especially important in our discussion of the isolated singularities of holomorphic functions, to be done next time.

We suppose that 0≤r<R≤∞, and define an annular region A by A={z in C with r<|z|<R}. (Here we will have our annulus and our Laurent series "centered" at 0. If you wish to center it at p, then |z–p| and (z–p) replace |z| and (z) in the following discussion.) We suppose that f is holomorphic in A. If z is in A and r is small, then the Cauchy Integral Formula implies that f(z)=(1/2Pi i)∫_|s–z|=r(f(s)/(s–z))ds. We now deduce the extremely useful Laurent expansion for f.

Deforming the integration contour continuously
The preliminary step is to distort the integration contour nicely. We "distort" |z–a|=r as shown in the picture. Notice that the distortions do not change the values of the line integrals since the function we are integrating is analytic in A\{z}.

The integrals over the line segments connecting the two circles cancel. We are left with integrals over two circles, one oriented positively (in the usual, counterclockwise, direction) and the other oriented negatively. Then Cauchy's Theorem implies that the integral of g over |s–z|=r is equal to the difference of two integrals, one over |s|=b and the other over |s|=a with r<a<|z|<b<R. Now we handle each of the integrals. I'm just going to outline the process -- it appears in virtually every complex analysis text. The opportunity for making sign errors and summation index errors arises frequently.

The outer integral
Well, consider [1/(2Π i)]∫_|s–z|=b(f(s)/(s–z))ds. We will "expand the Cauchy kernel". So:
We know |s|=b>|z|, so 1/(s–z)=((1/s))(1/(1–[z/s])). Now we use the geometric series, since |z/s|<1. The series will converge absolutely for each such s. We can get nice error estimates just as we did for the power series case. These estimates allow us to "interchange" integration and summation. I won't bother with the details of the estimates here.

The inner integral
Well, consider (1/2Π i)∫_|s–z|=a(f(s)/(s–z))ds. We will "expand the Cauchy kernel". So:
We know |s|=a<|z|, so 1/(s–z)=–((1/z))(1/(1–[s/z])). Now we use the geometric series, since |s/z|<1. The series will consider absolutely for each such s, and uniformly for s on |s–z|=a. Then we interchange integration and summation.

A difficulty: convergence of doubly infinite sums
Before we state the result, think a bit about how ∑_n=–∞^∞q_n should be defined. One possibility is to look at ∑_n=–N^Nq_n and require convergence as N→∞. This has some defects. For example, then the double series with q_n=–1 for n<0 and +1 for n>0 would converge (I guess to q₀). This series certainly would not converge absolutely, and the implication "absolute convergence implies convergence" is useful and we should preserve it. Also, if we used that definition, then relabeling n→n+1 might change the sum of the series. So the more accepted definition is as follows:

The series ∑_n=–∞^∞q_n converges and its sum is S if, given any real w>0, there is some positive integer N so that if m₁ and m₂ are both greater than N, then |∑_n=–m1^m₂ q_n–S|<w.

Then the double series will converge if and only if each of the half series (?), ∑_n=0^∞q_n and ∑_n=–∞⁰q_n, converges, and all of the expected statements will be correct.

Laurent Series Theorem
Suppose that 0≤r<R≤∞, and define an annular region A by A={z in C with r<|z|<R}. Further, suppose that f is analytic in A. Define the doubly infinite sequence of complex numbers {a_n} by a_n=∫_|s–z|=t(f(s)/sⁿ⁺¹)ds for any t with r<t<R. Then

• f(z)=∑_n=–∞^∞a_nzⁿ
• This series converges absolutely for every z in A and good error estimates can be gotten on any specific circle going around A.
• There is exactly one such series representing f in A.
• f can be written as a sum of f_in, analytic in {|z|<R}, and f_out, analytic in {|z|>r}. This decomposition is unique if also lim_z→∞f_out(z)=0.

Discussion of proof
Notice that in the formula for a_n, the integral ∫_|s–z|=t(f(s)/sⁿ⁺¹)ds, there seems to be a dependence on t. But Cauchy's Theorem declares that the result is the same for all t with r<t<R.

The existence of the series is obtained by following the path of expanding the Cauchy kernel twice, as mentioned before the statement of the theorem. If there is such a series, then the coefficients can be obtained by integrating f(s)/sⁿ⁺¹ around a circle, and interchanging sum and integral using the error estimates which I have refused to write! (Laziness or taste?)

Since Laurent series converge absolutely and we have good error estimates, then any of the standard manipulations of algebra (multiplication, addition, division, etc.) and calculus (differentiation, integration) are justifiable, and any sort of trickery which gets the series is justified.

f_in is the sum of the non-negative z powers, and f_out is the sum of the negative z powers. The limit at ∞ means that the constant term, if non-zero, goes with f_in.

Examples
I will compute some Laurent series for rational functions. For various reasons that will appear shortly, most other functions have series which rarely can be computed explicitly.

Let's consider the totally non-random rational function R(z)=(4z²–14z+8)/(z³–10z²+32z–32). Written that way I would have to work a bit to find series expansions. In fact, I got this by combining some simpler (partial) fractions. It is {1/(z–2)}+{3z/(z–4)²}. You can check this, hey, I won't do it here after all, this is Math 403 and we don't do the easy stuff. Maybe. I would like to find all possible Laurent series centered at i for R(z). I note that R(z) is analytic except for 2 and 4. And "clearly" there are three annuli (plural of annulus, surely) centered at i in which R(z) is analytic. These are

• |z–i|<sqrt(5) (which is the distance from i to 2; this annulus is actually a disc)
• sqrt(5)<|z–i|<sqrt(17) (the distance from i to 4)
• sqrt(17)<|z–i| (the only unbounded annulus considered here)

to be continued ...

The Cauchy estimates
Suppose f(z) is analytic in some connected open set U which contains the closed disc of z's with |z–a|≤r. Then for any non-negative integer, n, |f⁽ⁿ⁾(a)|≤n!M/rⁿ where M is the maximum of |f(z)| on |z|=r.

Proof Suppose C is the boundary of the circle |z–a|=r with the usual counterclockwise orientation. Then CIFD tells us that f(a)=[n!/{2Πi}]∫_C[f(z)/(z–a)ⁿ⁺¹]dz. The ML inequality gives us exactly the Cauchy estimate cited in the statement of the theorem because L is 2Πr, so the Π's cancel. Also |f(z)/(z–a)ⁿ⁺¹|≤M/rⁿ⁺¹ if M is as in the satement of the theorem. A power of r cancels so that we get the rⁿ in the bottom as the statement reads.

Probably the most interesting application of the Cauchy estimates is the following which is really totally novel.

Liouville's Theorem
A bounded entire function is constant.

This strange result is easy to prove. If f is entire (analytic everywhere in C), we'll show that f´(a)=0 if a is any complex number. So we know that |f(z)|≤M for all z. The Cauchy estimates, just for n=1, tell us that |f´(a)|≤1!M/r¹. But this is true for all r>0, and the only non-negative number |f´(a)| which can satisfy these inequalities is 0. So f´(a)=0 for all a's in C. The function is constant!

This is an amazing result, which a friend in grad school with me called the Louisville slugger because it is used many circumstances where it totally cleans up the situation -- there are no base-runners left. It has been difficult to explain this lovely pun to some of our non-U.S. grad students. Your homework has a further application, that if |f(z)| grows no faster than (Constant)|z|ⁿ, then f(z) must be a polynomial of degree≤n (so Liouville's Theorem is the case n=1).

Some counter(?)examples
With any new result, especially something so startling, exploring possible counterexamples is always useful.

• The function sin(z)
  As we can see from the picture so generously supplied, this function is not constant and is bounded by M=1. Explain that!
  It ain't bounded as an entire function. You can easily when z=i y and y is large positive, we've already shown that sin(z) is close to i e^y/2, an unbounded collection of numbers.
• The function 1/(1+z²)
  This function is also surely bounded, and the bound is, again, as displayed, 1. So it must be constant.
  f is analytic but not entire, since z=±i is not in the domain.
• The function 1/(1+|z|²)
  Now, since |z| is real and non-negative, I know that the domain of this function is all of C! Again, surely this is a constant function.
  But f is not analytic. I think |z| is not complex differentiable at any z, and actually |z|² is complex differentiable only when z=0, so |z|² isn't analytic anywhere. And don't call me Shirley.

Entire functions are very strange. For example, there is no non-constant entire function whose image is inside the unit disc (since such a function would certainly be bounded). And even more is true. Near the beginning of the course we worked with "simple" statements about complex numbers, modulus, complex conjugate, etc. You had a few homework assignments which looked like this:

When is w=(z–i)/(z+i) inside the unit circle? Of course this asks: when is ww<1? Let me run through the elementary solution again. Observe that all the steps are logical equivalences: they are reversible.

1. Since (z–i)/(z+i) is (z+i)/(z–i), this is the same as ((z–i)/(z+i))· ((z+i)/(z–i))<1.
2. This is the same as (z–i)(z+i)<(z+i)(z–i).
3. Expanding, this means zz–iz+iz+1 < zz+iz–iz+1 and we can clear away the zz and 1 from both sides.
4. Push everything to one side and divide by 2. The result is 0<iz–iz.
5. If z=x+i y, 0<iz–iz becomes exactly 0<i(x–i y)–i(x+i y) or –2i²y is positive or just y=Im(z)>0.

Any entire function whose values are in the upper half plane is a constant. For if it were not, then compose it with the mapping just discussed (it sometimes has the name Cayley transform) and that composition would have to be constant. But then compose this composition with the inverse mapping and the result is the original function, whose values are just one number. Thus the original function must be constant.

Any entire function whose values are in the union of the first, second, and third quadrants (those z's with 0<arg(z)<3Π/4) must be constant. Hey: compose the function with the mapping z→z^2/3 which can certainly be defined in that domain. It changes the 3/4 plane to the upper half plane (1-to-1 and onto since z^3/2 takes the half plane to the three quarters plane!) and then the composition is constant by the previous result. Etc.

These connected open sets make entire functions constant the range of such a function is are inside them.

What's going on? Well, you'll need to wait just a little bit more until we look at some geometry. First we have a major result which is historically significant to check. I will approach the result in a low-class (no: low-tech!) way.

Let me change gears and ask about a specific polynomial, let's say P(z)=z⁴–(3+i)z³+9.22z+14.5. Maybe P(z) has roots. If you do any sort of numerical work, an initial estimate is always useful. Is there some circle |z|=R where it is guaranteed that roots of P(z) must sit inside the circle?

How BIG or how small is |f(z)| on a typical circle? Well, certainly if P(z)=R≥1, then
    |P(z)|=|z⁴–(3+i)z³+9.22z+14.5|≤|z|⁴+|3+i||z|³+9.22|z|+14.5≤AR⁴
where A=max(1,|3+i|,9.22,14.5). Also,
    |P(z)|=|z⁴–(3+i)z³+9.22z+14.5|≥|z|⁴–|(3+i)z³+9.22z+14.5|. But |(3+i)z³+9.22z+14.5|≤BR³ if B=max(|3+i|,9.22,14.5).
In fact, if |z|=R≥2B, then |(3+i)z³+9.22z+14.5|≤(1/2)R⁴ so that |P(z)|≥R⁴–(1/2)R⁴=(1/2)R⁴.

Certainly P(z) has no roots if |z|>2B. These inequalities can be generalized.

If P(z)=zⁿ+∑_j=0^n–1a_jz^j, then there is K>0 and positive constants C₁ and C₂ so that for |z|≥K, C₁|z|ⁿ≤|P(z)|≤C₂|z|ⁿ.

The following "constants" will work: C₂ is just max(1,|a₀|,|a₁|,...,|a_n–1|) with K at least 1. C₁ can be 1/2 but K needs to be greater than 2C₂. So K=max(1,2C₂).

This growth lemma (which essentially says, in theoretical computer science language, |P(z)| has rate of growth |z|ⁿ as |z|→∞ (sometimes written |P(z)}=Ω(|z|ⁿ) when |z|→∞). This means |P(z)| can be estimated both above and below by multiples of |z|ⁿ.

The Fundamental Theorem of Algebra
If P(z)=zⁿ+∑_j=0^n–1a_jz^j, then there is at least one complex number a so that P(a)=0.

Once you know P(a)=0, then P(z)=(z–a)Q(z) where deg(Q)=n–1. So we can continue to find roots, and write P(z) as a product of linear factors.

Why is FTA true? Well, suppose that P(z) is never 0. Then define f(z)=1/P(z), an analytic function whose domain is all of C (entire!). I know that for |z|≥K, |f(z)|<1/{C₁|z|ⁿ} so |f(z)|≤1/{C₁Kⁿ} for those z's. But |f(z)| is continuous in |z|≤K so it has a bound, B, there (continuous functions are bounded on closed and bounded sets). Therefore f(z) is entire and bounded, hence (Liouville!) constant. But this is false since P(z) is not constant. So P(z) must have a root.

Mr. Cantave commented that this "proof by contradiction" wasn't totally useful -- it didn't come up with a candidate or approximate candidate for a solution. Well, there are other, constructive proofs. And, anyway, what does one expect in a math class. This seemed to be an opportune time to tell my favorite math joke, so I did.

JOKE Several people are in a hot-air balloon, trying to land over a fog-shrouded countryside at the end of a long day. The balloon dips down low and they see the ground faintly. Spotting a person, one of them calls down: "Where are we?" Some minutes later the wind is carrying them away and they hear faintly, "You're in a balloon!" One person in the balloon gondola says thoughtfully to the other, "It's so nice to get help from a mathematician." The other says, "How do you know that was a mathematician?" The first replies, "There are three reasons: it took a long time to get the answer, it was totally correct, and, finally, it was absolutely useless."

The proof I just presented to you is one of the traditional proofs, but, quick, before we discuss it, let me show you another verification, this one due to Professor Anton Schep of the University of South Carolina and published in 2009.

CIF applied to f(z)=1/P(z) at z=0 asserts that 2Πi/P(0)=∫_unit circlef(z)/{z–0} dz=∫_unit circle1/{zP(z)} dz. But we can deform the unit circle: this integral has the same value as the integral around a circle centered at 0 of radius R where R is large (the integrand is analytic between these curves -- it only fails to be analytic at 0). Then apply ML to ∫_|z|=R{1/[zP(z)]} dz. We get (2ΠR)/{RC₁Rⁿ} and this →0 as R→∞. But 2Πi/P(0) isn't 0, so again, a contradiction and we're done.

I think this is a sequence of lovely observations. It must be a nearly minimal verification of FTA. It was published in A Simple Complex Analysis and an Advanced Calculus Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 116, Jan 2009, 67-68 by Anton Schep. The article is available here.

FTA has a huge history and some special terminology. A field is algebraically closed if every polynomial with coefficients in the field has a root. A smaller field inside an algebraically closed field has the bigger field as its algebraic closure if all elements in the bigger field are needed to get roots of polynomials with coefficients in the smaller field. So we have just proved that C is algebraically closed, and that C is the algebraic closure of R. We need the elements of C which aren't "real" in order to guarantee that real quadratics have roots.

The history of FTA is complicated. There are books about FTA and its history. On the web, here is a comprehensive discussion of the history of the proofs. Another discussion is here. Probably FTA has hundreds of proofs. Many can be read here.
The doctoral dissertation of Gauss was about "a new proof of the Fundamental Theorem of Algebra". Many authorities believe that what he wrote then was actually the first credible proof of FTA, although its own rigor has been criticized. Discussions of Gauss's proofs (he gave 4 over his life!) are here and here.

There's also a more practical question of actually approximating the roots of a polynomial numerically: "constructive root finding". Anyone who computes things worries about roots of polynomials. That's been addressed in some of the references given above.

I further remark that FTA finally justifies (along with a lot of linear algebra!) the method of partial fractions, taught and sometimes learned (with much irritation!) in calc 2. Math is wonderful.

• If f is analytic on and inside a simple closed curve C, then the integral is 0. (Cauchy's Theorem).
• Not generally if f is not analytic, even if C is a simple closed curve. Hey, try z on the unit circle.
• Not generally even if f is analytic and C is not a closed curve. Try f(z)=z and C=[0,1].
• Not generally even if f is analytic on C and C is a simple closed curve. Try 1/z on the unit circle -- this is exactly what our noble ancestor Gauss means in this sentence extracted from the long quote on the back of the fourth homework assignment.

  "If, for example, we define log x via ∫{1/x}dx starting at x=1, then arrive at log x having gone around the point x=0 one or more times or not at all, every circuit adds the constant +2Π i or –2Π i; thus the fact that every number has multiple logarithms becomes quite clear."

Morera's Theorem
Suppose U is a connected open set and suppose that for all simple closed curves C in U, ∫_Cf(z) dz=0. Then f(z) is analytic.

Proof Well, if I want to verify that f(z) is analytic, then it is certainly enough to check that f(z) is analytic in an open disc inside U. To the right is a picture of such a disc. I take some point z₀ in the disc (a fixed point -- it doesn't matter which one) and I take any path P from z₀ to a general point z. I guess P should be piecewise differentiable, etc. Then I define F(z) to be ∫_Pf(z) dz. Here physics people might recognize that choosing z₀ is choosing a "ground state" again. The value of F(z) does not depend on the choice of P, since the "integral vanishing" hypothesis of Morera's Theorem implies that f(z) dz is path independent.

So now I'll verify that F(z) is (complex) differentiable, and that F´(z)=f(z). The astute student (even a barely awake student!) will notice that this is the third time I've done this proof. What the heck -- this is such fun.

We can take a path as illustrated (ending with a horizontal line segment) to compute F(z). Let me just look at the last part carefully (with u(x,y) and v(x,y) being the real and imaginary parts of f(z)):
    F(z)=∫_{vertical piece}f(z) dz+∫_x0^xu(t,y)+i v(t,y) dt.
Here for the second piece, the horizontal part, I have parameterized z=t+i y₀ with t going from x₀ to x, z₀=x₀+i y₀, and z₀=x+i y, Notice if I change x, the vertical piece of the integral doesn't change. I can differentiate the formula with respect to x using FTC of calc 1. The derivative is u(x,y)+i v(x,y). By the way, if F(z)=U(x,y)+i V(x,y) is F's decomposition into real and imaginary parts, we have just shown that U_x=u and V_x=v.

Here is the other picture we need (of course!). Now I've written a path for which it will be easy to compute the y derivative of F(z). So:
    F(z)=∫_{vertical piece}f(z) dz+∫_y0^y[u(x,s)+i v(x,s)]i ds.
Here in the second (vertical) part I parameterized by z=x+i s so dz=i ds (don't forget the i!). s goes from y₀ to y. Taking a y derivative uses FTC again, and the result on the right is [u(x,y)+i v(x,y)]i which is –v(x,y)+i u(x,y). If F(z) is U(x,y)+i V(x,y), then the left-hand side differentiates to U_y+i V_y.

Let me summarize. We know this: U_x=u and V_x=v; U_y=–v and V_y=u. What does this mean? Well, we see that U_x=V_y and U_y=–V_x. Therefore the real and imaginary parts of F(z) satisfy the Cauchy-Riemann equations so we know that F(z) is analytic. And the specific equations tell us that F´=f. But the derivative of an analytic function is analytic, so f is also analytic. We're done.

This is an amazing result because we got differentiability out of the "machine" without any apparent effort (and we also got an antiderivative for f(z), but that's not too helpful in practice!). I will show how to use this in a second in a surprising way, but I should observe that Morera's Theorem joins the previous catalog of equivalent statements. I also "split" the power series statement into two statements, one of which looks much "stronger" than the other. They are equivalent to all the other statements.

A magical list A remarkable fact which I've hinted at repeatedly but not explicified (not a word!) is that a number of different characterizations of functions are logically equivalent. This collection of results leads to a very wide variety of tools to attack all sorts of computational and theoretical problems. Suppose U is an open connected set and f:U→C is a function. Here is what we know now: The definition f has a complex derivative, f´(z), at each point z of U (this is defined to be limh→0{f(z+h)–f(z)}/h) and f´ is continuous. (Goursat showed that the requirement of continuity can be dropped but I know no natural situation where the continuity of f´ doesn't happen somehow automatically.) Power series I f has a power series expansion valid near every point z0 of U: that is, given such a point, there is a positive number r which generally depends on z0 and a sequence of complex constants {an}n≥0 so that the series ∑n=0&infinan(z–z0)n converges for |z–z0|<r with its sum equal to f(z). Power series II f has a power series expansion valid near every point z0 of U: that is, given such a point, there is a positive number r which generally depends on z0 and a sequence of complex constants {an}n≥0 so that the series ∑n=0&infinan(z–z0)n converges for |z–z0|<r with its sum equal to f(z). The positive number r is at least the distance from z0 to the boundary of U (because we can push out the circle in the proof of the CIFD until it "hits" the edge of U). Differentiation f has a continuous derivative f´ and this derivative also has a continuous derivative. More derivatives f has any number of continuous derivatives you may want (as long as the number is at least 1 [sigh]). Cauchy-Riemann If we write f as u+i v where u and v are real functions, then u and v have continuous partial derivatives and satisfy the two partial differential equations ux=vy and uy=–vx. Harmonic functions If the real part of f is u, then u is harmonic, Δu=uxx+uyy=0, and the imaginary part of f (that's v) is also harmonic and these two harmonic functions are harmonic conjugates of each other. Morera's Theorem f(z) is continuous, and for all closed curves C in U, ∫Cf(z) dz=0. A geometric characterization This is called conformality and we'll get to it in a few weeks, after we do a bunch of applications of what's already here.

Let me show you how to use this with a complicated computational example. But first, a digression (!?) to calc 2. We have certainly exploited many times the facts about geometric series which are reviewed in calc 2 (after being learned, maybe, in either high school or junior high). Another family of series is studied in calc 2.

p-series
The p-series is the series 1+1/2^p+1/2^p+1/3^p+1/4^p+... (etc.). You may remember that this series converges when p>1. It has to converge absolutely since the series consists of positive terms. How fast does it converge? Let me recall some of calc 2, and let me temporarily concentrate on p=2. Now ∑_n=1^∞1/n²=∑_n=1^N1/n²+∑_n=N+1^∞1/n². How big is the "infinite tail", ∑_n=N+1^∞1/n²?
The standard calc 2 technique for estimating this sum is to compare it to an integral. I always get confused about this, but a picture helps. The boxes are under the curve y=1/x², and so the infinite tail is less than ∫_N^∞[1/x²]dx. But I can find exact antiderivatives of powers, and then plug things in. The value of this improper integral is 1/N. So the tail ∑_n=N+1^∞1/n² is less than 1/N. By the way, more numerical work shows that this is about as good as you can expect. The tail can't be estimated much better than this asymptotically.

What if I wanted to know for which N the partial sum will be within .001 of the value of the whole sum? Well, .001 is 1/1,000 so I would need a thousand terms. By contrast, if we look at the geometric series ∑_n=1^∞1/2ⁿ, the corresponding tail ∑_n=N+1^∞1/2ⁿ is exactly 1/(2^N). To get accuracy of .001, I would just need N=10. Or, another way, if I took N=100, I'd get accuracy of ... uhhh ... less than 8·10^–31, which seems small. These p-series converge much slower than geometric series.

What if I vary the p? If I make p larger than 2, then the terms get smaller (1/n^p decreases as the positive real number p increases). So in fact I get an error of <.001 by taking at least 1,000 terms for any p-series with p≥2.

Complex p-series?
Remember that a^b=e^b log(a) with log(a)=ln(a)+i arg(a). Here I will take a to be a positive integer n, and b will be z=x+i y. Also since n is a positive real, let me just use the principal branch of log, so Arg(n) is 0. Now n^z=e^{(x+i y)ln(n)}=e^x ln(n)e^(i y)ln(n). The modulus of n^p is exactly e^x ln(n)=n^x since e^{i (any real)} has modulus 1. So |1/n^z| is just e^{–ln(n)Re(z)}=1/n^x. Therefore the modulus of ∑_n=1^∞1/n^z will converge absolutely if x=Re(z)>1 because I can compare the terms to terms of the usual p-series with p=Re(z).

The series ∑_n=1^∞1/n^z converges absolutely and therefore converges in the open half-plane Re(z)>1. I will temporarily call this sum Q(z). What sort of function is it? In fact, I claim this sum is actually an analytic function. Please note that verification of this by, say, using the definition is not so easy. The examples I showed you of Fourier series where the sum of the derivatives is not the derivative of the sum should perhaps suggest that. This series is not a power series, and it is much worse (converges much slower) than such a series. We should be careful even though each piece, 1/n^z=e^–ln(n)z, is easily seen to be analytic (just differentiate with the Chain Rule!). So I'll be a bit cute here and use Morera's Theorem.

If I took, say, z's with Re(z)≥2, I know that Q(z)=∑_n=1^N1/n^z+Error(N) where Error(N) is a complex number and |Error(N)|<1/N. This is because any such series can be overestimated by the ordinary p-series with p=2 and the infinite tail can be overestimated by the same estimate of the usual infinite tail. Now suppose C is a simple closed curve of length L in the half plane Re(z)≥2. Then:
    ∫_CQ(z) dz=∫_C∑_n=1^N1/n^z dz+∫_Cinfinite tail dz
Each of the pieces 1/n^z in the finite sum is analytic and therefore (Cauchy's Theorem) ∫_C∑_n=1^N1/n^z dz=0. What about ∫_Cinfinite tail dz? We estimate its modulus using ML. The M is 1/N so this piece has modulus less than L/N. But we can do this for any N. This means that the modulus of the integral we started with, ∫_CQ(z) dz, is less than L/N for any positive integer N. There is only one non-negative number like that: 0. So the integral over C of Q(z) is 0.

The p=2 restriction is sort of silly, and we could do any p>1 similarly. The function Q(z) is actually analytic in Re(z)>1.

What's going on?
I haven't told you the REAL NAME of this function.

The Cauchy Integral Formula for Derivatives Suppose R is a connected open set, C is a simple closed curve whose inside is contained entirely in R, p is a point inside C, and f is analytic in R. Then for any positive integer n, f is n times differentiable, and f(n)(p)={n!/[2Π i]}∫C[f(z)]/[(z–p)n+1]dz. I think I proved this only for a circle which has p inside it, but any C described in the theorem can be deformed to such a circle, so the result is true also. I will try to substantiate this by drawing an appropriate picture such as the one to the right, which I've already drawn once. For this change, though, just reverse the arrows: make the circle into C. Notice that the integrand, [f(z)]/[(z–p)n+1], is analytic in all of the region except for p so this deformation does not change the value of the integral.

A strange estimate on the size of derivatives
If you really believe in the CIF for derivatives, then very strange things happen. Let me show you. Suppose we take C as above inside a connected open set R, and the inside of C is entirely inside R, as shown. Suppose all of our p's are in a set K and all the p's are at a distance at least d to the curve C. So the setup is sort of what is pictured to the right. Now I know that if p is any point in K, f⁽ⁿ⁾(p)={n!/[2Π i]}∫_C[f(z)]/[(z–p)ⁿ⁺¹]dz. Let me apply the ML inequality in a very direct way. The length of C will be L. And I want to estimate the modulus of the integrand, [f(z)]/[(z–p)ⁿ⁺¹]. Let me label the maximum of the modulus of f(z) on C with M. I need to underestimate |z–p| where z is on C and p is in K. Well, certainly one underestimate is d. I am not going to get anything very precise because I haven't given you very much specific information. We get this sort of result:
       If p is in K, then |f⁽ⁿ⁾(p)|≤{n!/[2Π]}CM/dⁿ⁺¹.
Now I don't want to beat this inequality to death (that will occur next time, when you'll see it is even more weird than it seems today) but notice the M on the right-hand ("upper") side. If M gets very very small, so the modulus of the function is tiny, this inequality declares that the derivatives inside must also get very very small. So a small function (in the sense of modulus) implies small derivative (in the sense of modulus). Derivatives sort of control the amount of wiggling in a function, and this means that very small functions can't wiggle very much. Understand?

A counter(?)example(?)
The best examples I know of in math are the ones which force you to reconsider your assertions, or, at least, to look very closely at them. Look now at the sequence of functions {sin(n²x)/n} where n is a positive integer. Certainly these functions are bounded (their values are in the interval [–1/n,1/n]). As n→∞, certainly the sizes (maximum heights of these functions) →0. But, wait: their derivatives are n cos(n²x), and certainly these functions, the first derivatives, get really really big as n→∞ (the values occupy all of [–n,n]). There is a great deal of wiggling going on. (That's 3 certainly's, so everything should be totally clear [NO!].)

What is happening?

Of course the answer is that the complex sine is not bounded at all away from the real axis. We should really look at the complex picture. We need a closed curve around a piece of the real axis. To the right is shown such a curve, C, around a chunk of the real axis. Notice that the curve crosses the imaginary axis at two places.

Remember that if y is real, sin(i y)=i sinh(y), and sinh(n²y) is exactly {e^n2y–e^–n2y}/2. The second piece if y is positive →0 rapidly as n→∞. But for fixed non-zero positive y, sinh(n²y) is like e^n2y/2 for n large. It grows really, really fast! So both the example and the previous estimate (which will return shortly in a form called the Cauchy estimates) are correct but they essentially have little to do with one another. What happens on the real axis to an analytic function is not a complete qualitative description of the behavior of the function.

A magical list A remarkable fact which I've hinted at repeatedly but not explicified (not a word!) is that a number of different characterizations of functions are logically equivalent. This collection of results leads to a very wide variety of tools to attack all sorts of computational and theoretical problems. Suppose U is an open connected set and f:U→C is a function. Here is what we know now: The definition f has a complex derivative, f´(z), at each point z of U (this is defined to be limh→0{f(z+h)–f(z)}/h) and f´ is continuous. (Goursat showed that the requirement of continuity can be dropped but I know no natural situation where the continuity of f´ doesn't happen somehow automatically.) Power series f has a power series expansion valid near every point z0 of U: that is, given such a point, there is a positive number r which generally depends on z0 and a sequence of complex constants {an}n≥0 so that the series ∑n=0&infinan(z–z0)n converges for |z–z0|<r with its sum equal to f(z). Differentiation f has a continuous derivative f´ and this derivative also has a continuous derivative. More derivatives f has any number of continuous derivatives you may want (as long as the number is at least 1 [sigh]). Cauchy-Riemann If we write f as u+i v where u and v are real functions, then u and v have continuous partial derivatives and satisfy the two partial differential equations ux=vy and uy=–vx. Harmonic functions If the real part of f is u, then u is harmonic, Δu=uxx+uyy=0, and the imaginary part of f (that's v) is also harmonic and these two harmonic functions are harmonic conjugates of each other. Morera's Theorem I'll discuss this very shortly but it links path independence to all the other characterizations. A geometric characterization This is called conformality and we'll get to it in a few weeks, after we do a bunch of applications of what's already here.

I will try to persuade people that we actually know that all of these characterizations are equivalent already, and that we know even more (e.g, the Cauchy integral formula for derivatives ties together integrals of f and the coefficients of the various power series expansions of f). Let me show you some very slick applications of these equivalent statements (just a few -- most of the remainder of the course will be such applications!).

Steady state heat flows aren't kinky
In physics and certain other areas of applications, people study harmonic functions as a key part of the mathematical model of their discipline. For example, a solution u of Δu=0 might be a steady-state heat flow, or it might be an electromagnetic field away from a charge or ... lots of things. A very similar differential equation is the Wave Equation, which is u_xx–u_yy=0. The phenomena which are modeled by the Wave Equation include things like sound and other vibrations. The solutions of the Wave Equation characteristically (a bit of a joke, that word, if you've had a PDE course!) have breaks or shocks -- the mathematical counterpart of what we can see visibly in vibrations and other phenomena. But if we change – to +, here is what happens:
Suppose we have u_xx+u_yy=0. Pick any point p and any disc centered at p in the domain of u. In lecture #10 on February 22, we brute force "created" a harmonic conjugate v for u just in that disc. The proof was yucky, but the fact was verified. So there is v so that u+i v is analytic. I am not declaring that ALL of u has a harmonic conjugate, since in the same lecture I gave an example of a u without a harmonic conjugate. But "locally" I can always find one. Now f=u+i v can be differentiated any number of times. So therefore u can be differentiated also. Therefore, any harmonic function, assumed only twice differentiable in its definition, is actually infinitely often differentiable! Heat flow has no shocks.

This is not an obvious conclusion.

But there really are kinky (very weird) functions!
Almost all of the ideas covered in this course are mathematics which was created and maybe nearly perfected in or before the 19^th century. So perhaps I should look a little bit forward, and tell you what wasn't well known to professors in 1850. The specific example I'm about to discuss also shows some of the power in what we have already proved. So here is (nominally!) an example from calc 1.

Consider the function f:R→R defined piecewise by f(x)=0 for x≤0 and f(x)=e^–1/x for x>0. A Maple graph of this function is shown to the right.

The difficulties in this function occur at/near x=0. For x>0, the value e^–1/x is exp(a negative number). Therefore, if x>0, 0<f(x)<1. In fact (we are "in" calc 1 mode) for x>0, f´(x)=e^–1/x(1/x²). This function is positive for all x>0, so f(x) is increasing for x>0. Also, note that lim_x→∞e^–1/x=1 because when x is large positive, –1/x is small positive, so exp of that is close to 1. Also, lim_x→0⁺e^–1/x=0. Why? If x is small and positive, then –1/x is large and negative, and exp of that is close to 0. In fact, we have just verified that f(x) is a continuous function for all values of x. This is "obvious" for x<0 and for x>0 (there f(x) is given by a nice formula). The only value we really need to check is at x=0: we just did this.

What may not be clear is differentiability of f(x). If x<0, sure it is, and I bet that f´(x) is 0 for those x's. If x>0, I know that f´(x) exists and is actually given by the wonderful formula f´(x)=e^–1/x(1/x²). Again, the difficult "point" is x=0. We need to study lim_h→0[f(0+h)–f(0)]/h. Well, if h<0 the top is (0–0)/h which is 0. So we need to check that lim_h→0⁺[f(0+h)–f(0)]/h exists and is 0. Now f(0)=0 (that's in the darn definition!). And for h>0, f(0+h) is e^–1/h. So we need to "compute" lim_h→0⁺e^–1/h/h. A few people remembered the contortions (?) which are needed to compute this. Here is how to do it.

To compute lim_h→0⁺e^–1/h/h, change variables: let's call 1/h the new variable w. Then what does "h→0⁺" turn into? If h is small and positive, then w is large and positive. And e^–1/h/h becomes e^–ww, so that the limit turns into lim_w→∞e^–ww=lim_w→∞w/e^w. This limit, finally, can be computed with L'Hopital, and we differentiate the top and bottom to get the limit of 1/e^w as w→∞. This limit is easily equal to 0.

Wow. We have just used only a sort of simple-minded analysis to, in a "brute force" way, show that f is differentiable at x=0, and that f´(0)=0.

So we have a description of f´. Here it is: f´(x)=0 for x≤0 and f´(x)=e^–1/x(1/x²). What is more amazing is that this function is, in turn, differentiable. For x<0, of course f´´(x)=0. For x>0, if I use the Chain Rule and the Product Rule correctly (which I didn't in class!) then f´´(x)=e^–1/x({1/x⁴}–{2/x³}). What happens at x=0? Again, we think of a two-sided limit, and the left-hand limit is just 0. Therefore we need to consider lim_h→0⁺{f´(0+h)–f´(0)}/h. This becomes lim_h→0⁺e^–1/h(1/h³) using the known value of f´(0) (that's 0) and the formula for f´(x) when x>0. The same trick as before (1/h becomes w etc.) turns this limit into lim_w→∞w³e^–w. Well, we think of w³/e^w and use L'Hop three times. Etc.

Etc. indeed!
It turns out (o.k., darn it, a proof using mathematical induction) that f can be differentiated infinitely often (any number of times) and that for any positive integer n, f⁽ⁿ⁾(0) is 0. The graph of y=f(x) is flat, really flat, infinitely flat (!) at the origin. This is a bit distressing because of the following statements.

• The Taylor series for f centered at x=0 is ... 0. (Yes it is! All of the coefficients are 0.)
• The radius of convergence of the Taylor series of f centered at x=0 is ∞ (well, darn it, the series does converge for every x).
• There is no interval centered at x=0 for which f(x) is equal to its Taylor series centered at 0.

Deforming curves but not changing the value of line integrals
Here is a more complicated corollary of Cauchy's Theorem. It will allow us to compute many more integrals by comparing them with one another.

Corollary
Suppose that f(z) is analytic in a region R, and that C₁ and C₂ are simple closed curves in R. Suppose also that the curves can be continuously deformed, one into the other, all inside the region R. Then ∫_C1f(z) dz=∫_C2f(z) dz.
About the picture The picture attempts to illustrate the situation described in the corollary. The region R is a connected open set. The green arrows show how the curve C₂ is deformed into C₁ through part of R (but not through any of the holes!). If we picked a point w₂ on C₂, the deformation would move it along a curve to a point w₁ on C₁: let's call that curve B.
Proof Look at this curve, which is a sum of curves: start at w₂ and move along C₂ until we get to w₂ again. Then move on B until we arrive at w₁. Now move backwards on C₁, returning to w₁. Finish by moving backwards on B. The curve described is C₂+B–C₁–B. We apply Cauchy's Theorem to that curve, because f(z) is analytic on and inside the curve. The integral is 0, but that means 0=∫_{C2+B–C1–B}f(z) dz=∫_C2f(z) dz+∫_Bf(z) dz–∫_C1f(z) dz–∫_Bf(z) dz, so (since the B integrals cancel) we get ∫_C1f(z) dz=∫_C2f(z) dz.
Comment Here is some (negative) discussion about this "proof". What do I mean precisely by "deformation"? I haven't described this. Well, the notion can be described precisely, but I won't do it in this course, and in any situation I use this result in this course will be one where a picture verifying the proof can be explicitly drawn. Another objection can be applying Cauchy's Theorem to the curve C₂+B–C₁–B. This really isn't a simple closed curve. Again, this is true, but it is a limit of simple closed curves (think of twin B's slightly separated, with accompanying twin w₂'s and w₁'s). Each of those curves would be simple closed curves, and each would have integral equal to 0. The limiting curve with limiting integral would then also have integral equal to 0. The idea is what I'm showing here, and I am willing to admit that I've given up some precision!

The one non-zero integral we need ... We know that ∫|z|=1(1/z)dz=2Π i. This is a vital fact. (You can compute this directly using z=eit etc.)

A strange choice of function
What I'm describing is a somewhat complicated setup. I don't think I'll make it better by declaring that the whole procedure is one which is used repeatedly in many areas of applied mathematics, engineering, and physics. I will try, though, to describe the simplest incarnation of the situation. So here are the ingredients:

• A connected open set R and a function f(z) which is analytic in R.
• A simple closed curve C whose inside is contained entirely in R.
• A random (!) point p inside C.

The Cauchy Integral Formula
Suppose R is a connected open set, C is a simple closed curve whose inside is contained entirely in R, p is a point inside C, and f is analytic in R. Then f(p)={1/[2Π i]}∫_C[f(z)]/[z–p]dz.
Comment This is a fundamental and remarkable result. Differentiability, that is, complex differentiability, has the consequence that knowing the values of f on C determines the values of f inside C. We need to sort of "average" the values properly with weights determined by 1/[z–p] but everything is determined by the boundary values. This has no counterpart in real calculus. Certainly a differentiable function's value inside an interval is not determined by what happens at the endpoints of the interval.

An example and exercise: a trig integral
I asked Maple to compute ∫₀^2Π1/[3+sin(t)]dt and the response I got was Πsqrt(2)/2. Let me show you one way this can be done. I will use complex analysis and the Cauchy integral formula. There are short cuts we will develop in the next few weeks, but I want to present you with a cute direct application of CIF (the Cauchy integral formula).

I will "reverse engineer" the transformation which we've done many times in this course, where we started with a line integral and then, by parameterization, got a standard definite integral. Here we will start with ∫₀^2Π1/[3+sin(t)]dt. Certainly the bounds 0 and 2Π on the definite integral suggest "the unit circle" to me, so z=e^it and 0≤t≤2Π. Then dz=i e^itdt so that dz/(i e^it)=dt and dz/(i z)=dt. But I need to know how to change sin(t) into "z-land". Here my knowledge of power series helps, and I remember that sin(t)=[e^it–e^–it]/2i=[z–(1/z)]/2i. Now the integral we started with becomes:
    ∫_|z|=1(1/[3+{z–1/z}/2i]((dz/iz)=∫_|z|=1(2/(z²+6i z–1))dz.

That's if I did the algebraic "massage" of the integrand correctly. Now where is the integrand analytic? Well, everywhere except at the roots of z²+6i z–1. The Quadratic Formula applies here (it is a problem in one of the first sections of the book, but, actually, the Quadratic Formula applies in any field). The roots are: [–6i±sqrt{(–6i)²+4}]/2. Under the square root we have (–6i)²+4=–36+4=–32. We divide everything by 2 (by 4 under the square root!) and get as roots (–6±2sqrt(2))i. Well, r₁=(–6–2sqrt(2)) i is inside the unit circle and r₂=(–6+2sqrt(2)) i is outside the unit circle. But the integral we have is ∫2/[(z–r¹)(z–r²)]dz. This is exactly CIF with f(z)=2/(z–r₂) and p=r₁, and therefore the value of the integral is 2Πi f(r₁) which is 4Πi/(r₁–r₂) which turns out to be ... 4Πi/(4sqrt(2)i), the same as Maple's Πsqrt(2)/2.

Expanding the Cauchy kernel
The Cauchy kernel is the function 1/[z–p]. It can be manipulated algebraically in a variety of ways to give properties of analytic functions. I'll do a basic example of this manipulation here. Later we will do others.
I will suppose f(z) is analytic in some big disc centered at 0. For my simple closed curve I will take a circle of radius R centered at 0. The radius will be smaller than the radius of the "big disc". Also p will be some point inside the circle of radius R. Then I know f(p)={1/[2Π i]}∫_|z|=R[f(z)]/[z–p]dz. Consider 1/[z–p]. Since z is on |z|=R and p is inside the circle, |p|<R=|z|. |p/z| is less than 1. So do this:
1/[z–p]=[1/z]/[1–{p/z}]: this is supposed to make you think of the formula for the sum of a geometric series, a/[1–r], since we have arranged for r to be p/z, a complex number of modulus less than 1. We could write the whole infinite series, or just stop after a few terms. Let me show you.
[1/z]/[1–{p/z}]=∑_n=0^∞pⁿ/zⁿ⁺¹=∑_n=0^Npⁿ/zⁿ⁺¹+∑_n=N+1^∞pⁿ/zⁿ⁺¹=&sum_n=0^Npⁿ/zⁿ⁺¹+[p^N+1/z^N+2]/[1–{p/z}].
Here the ERROR, the difference between the partial sum and the whole series, is [p^N+1/z^N+2]/[1–{p/z}]. The modulus of this is at most (for any z on the circle) [|p|^N+1/R^N+2]/[1–{|p|/R}] and since |p|/R<1, ERROR→0 as N increases.
Now let's stuff the preceding sum into the integral:
∫_|z|=R[f(z)]/[z–p]dz=∫_|z|=R[f(z)](∑_n=0^Npⁿ/zⁿ⁺¹+ERROR)dz
The finite sum is really ∑_n=0^N∫_|z|=R[f(z)] pⁿ/zⁿ⁺¹dz=∑_n=0^N(∫_|z|=R[f(z)]/zⁿ⁺¹dz)pⁿ.
The other term can be estimated by ML: the length is 2Π R while the M is the maximum of |f| on the circle |z|=R multiplied by [|p|^N+1/R^N+2]/[1–{|p|/R}]. Since |p|/R<1, this will go to 0 as N increases. But put everything together. We have just shown that a certain infinite series converges, and since we know the Cauchy integral formula, we also know the sum of the infinite series.

f(p)=∑_n=0^∞a_npⁿ where a_n=[1/{2Π i}]∫_|z|=R[f(z)]/zⁿ⁺¹dz.

Complex differentiability implies power series
Here is a general statement of the preceding result.
Theorem Suppose f(z) is analytic in a connected open set which includes some disc centered at a point z₀. Then there are complex numbers {a_n}_n=0^∞ so that the series ∑_n=0^∞a_n(z–z₀)ⁿ converges at least for all z in that disc centered at z₀. Additionally, the coefficients a_n are given by the formula a_n=[1/{2Π i}]∫_|z–z0|=R[f(w)]/(w–z₀)ⁿ⁺¹dw where R is any positive number less than the radius of the disc.

Consequences
First, since power series can be differentiated inside their radius of convergence, we see that a complex differentiable function has a derivative which is itself a complex differentiable function! This is totally unlike real calculus. For example, we know that |x| can't be differentiated at 0. It has an antiderivative, let's call it h(x), given piecewise by x²/2 for x>=0 and by –x²/2 for z<0. The function h(x) has a derivative (just |x|) and it does not have a second derivative. In the complex "realm", functions always have more derivatives! There aren't any problems like |x|. One derivative implies having infinitely many derivatives!

Second, we saw when we studied power series that a function can have only one power series representation centered at a point. That is, if f(z)=∑_n=0^∞a_n(z–z₀)ⁿ and f(z)=∑_n=0^∞b_n(z–z₀)ⁿ, both converging near z₀, then each a_n is equal to the corresponding b_n. In fact, since we can differentiate power series, the power series must be a Taylor series: a_n=(1/n!)f⁽ⁿ⁾(z₀). But we can compare the two descriptions of the coefficients a_n we have, and then we can deduce the following result.

The Cauchy Integral Formula for Derivatives
Suppose R is a connected open set, C is a simple closed curve whose inside is contained entirely in R, p is a point inside C, and f is analytic in R. Then for any positive integer n, f is n times differentiable, and f⁽ⁿ⁾(p)={n!/[2Π i]}∫_C[f(z)]/[(z–p)ⁿ⁺¹]dz.

I think I proved this only for a circle which has p inside it, but any C described in the theorem can be deformed to such a circle, so the result is true also.

Here are graphs of the partial sums of some trigonometric series.

If w is real, then |sin w|≤1 so the infinite series ∑_j=1^∞sin(2^jx)/j⁵ converges for all x because it converges absolutely. An easy estimate with an improper integral (∑_j=51^∞(1/j⁵)<∫_j=50^∞dx/x⁵) shows that the partial sum of the first 50 terms is within .00000004 (yes!) of the whole sum.

Here is a Maple graph of the partial sum ∑_j=1⁵⁰sin(2^jx)/j⁵ on the interval [0,2Π]. Humans would not notice any discrepancy as small as what was just written. I think the graph of the entire series would look the same -- for the scale displayed!

And, by the way, the graph is not the graph of sin(2x) (I worried about that also!). A graph of the difference is shown to the left, with vertical range about [–.03,.03]

Now here is a Maple graph of ∑_j=1⁵⁰2^jcos(2^jx)/j⁵, a partial sum of a much stranger trig series. I just differentiated each term in the partial sum of the first series and added the results. Notice that the graph is emphatically not nice!.

The interval of x's is again [0,2Π] but Maple has scaled the vertical axis. It is about [–5·10⁶,8·10⁶]. The graph isn't reliable here because of the difficulties of plotting a function with such wiggly behavior.

The series ∑_j=1^∞2^jcos(2^jx)/j⁵ probably doesn't converge very often. For example, suppose x=Π/2^k for some positive integer, k. Then as soon as the index of summation, j, passes k, the cosine function is evaluated at a multiple of 2Π so the terms of the series are 2^j/j⁵, which does not →0, Many other x's have similar or even more bizarre properties. Therefore the term-by-term derivative of the original series does not converge.

I am not asserting that the original series is not differentiable (its sum is actually yet another "special function" which Maple identifies and can differentiate). I do assert that the simple generalization of "The derivative of a sum is the sum of the derivatives" to an infinite series is false in this case. Power series are very nice. Trigonometric series are not.

So here is what some of the niceness causes: a few consequences of the third statement about power series:

Suppose R≠0. Then the sum of the series &sum_j=0^∞a_j(z–z₀)^j, inside the radius of convergence (that is, for |z–z₀|<R). is a complex differentiable function whose derivative is the sum of the series &sum_j=0^∞j a_j(z–z₀)^j–1 which has the same radius of convergence.

Suppose f(z)=&sum_j=0^∞a_j(z–z₀)^j. Then f(z₀)=a₀ because all of the other terms drop out (z₀–z₀=0).

Let's differentiate: f´(z)=&sum_j=0^∞j a_j(z–z₀)^j–1. Now "plug in" z₀. The result is f´(z₀)=a₁. The j>1 terms drop out (z₀–z₀=0 again) and the j<1 term drops out (multiplication by j).

Differentiate again: f´´(z)=&sum_j=0^∞j(j–1) a_j(z–z₀)^j–2 and plug in again. So f´´(z₀)=2a₁. The j>2 terms vanish as before, the j<2 terms are 0 (because of the j(j–1) factor), and the 2 comes from j(j–1) instantiated for j=2.

Repeated differentiations and "plug ins" show that f^(j)(z₀)=j!a_j, so that a_j=f^(j)(z₀)/j!. Therefore:

• Every convergent power series is a Taylor series for its sum.
• A function which is equal to a power series with a non-zero radius of convergence centered at z₀ has a unique such series, so any series centered at z₀ must be the same.

Today we leave real calculus behind. What will occur is quite novel. There are various methods for deducing the results I'll show you but here I'll stick with basically classical techniques, about 150 years old. A useful reason to do this is that this way you'll see methods which have been generalized and used in many fields of pure and applied mathematics.

Almost all integrals are 0
Well, let me repeat a version of Cauchy's Theorem which relies totally on Green's Theorem for its verification. This was first discussed in this course in the snow make-up lecture.

Cauchy's Theorem Suppose f(z) is analytic on and inside a simple closed curve C. Then ∫_Cf(z) dz=0.
Proof Suppose that f(z)=u(x,y)+iv(x,y) where u and v are real-valued functions (the real and imaginary parts of f). Then f(z) dz=[u(x,y)+iv(x,y)][dx+idy]=[u(x,y)dx–v(x,y)dy]+i[v(x,y)dx+u(x,y)dy]. The statement of Green's Theorem we will use is that ∫_Cpdx+qdy=∫∫_Rq_x–p_ydA where R is the region bounded by C, dA means area (dxdy, if you would rather) and both p and q are continuously differentiable everywhere in C and R (even one point with a problem might make the equation invalid!).
Let's look at ∫_C[u(x,y)dx–v(x,y)dy] with p=u and q=–v. Then q_x–p_y is –v_x–u_y. But one of the Cauchy-Riemann equations is u_y=–v_x, so the integrand in the double integral over R is 0.
Let's look at ∫_C[v(x,y)dx+u(x,y)dy] with p=v and q=u. Then q_x–p_y is u_x–v_y. The other Cauchy-Riemann equation is u_x=v_y, so the integrand in this use of Green's Theorem is 0 also.
Both the real and imaginary parts are 0, so the integral is 0. We're done.

A note about Goursat
The key hypothesis in Cauchy's Theorem is that u and v are continuously differentiable and satisfy the Cauchy-Riemann equations in the appropriate region. But Goursat, late in the 19^th century, showed that that if f´(z) exists, then the conclusion of Cauchy's Theorem follows. The continuity of the derivative is unnecessary. The proof is in the text, and is quite clever. I won't give it here, because to my knowledge the techniques of the proof are rarely used in most of complex analysis.

An example and exercise: Fresnel
This is problem 13 in section 2.3, and is a very neat direct application of Cauchy's Theorem. The problem is to compoute Fresnel_C=∫₀^∞sin(t²)dt and Fresnel_S=∫₀^∞cos(t²)dt, the "Fresnel integrals" (more about them in a second). This is a standard use of complex analysis. There are other methods of computing this integral but this is the customary may.

It may be difficult to convince yourself that the integrals even exist before we compute them. But look at the graph of sin(t²) from 0 to 10. The alternating "bumps", getting narrower and narrower, clearly (?) almost cancel each other out.

Let's compute the integral of f(z)=e^iz2 around the simple closed curve from 0 to R (R large, real, positive) then around a circular arc to (1+i)R/sqrt(2), and then back to the origin. Since f(z) is entire, the integral is 0 for any R. I'll call I_R the line segment along the real axis, C_R the circular arc, and J_R the line segment "returning" to 0. Cuachy's Theorem implies that ∫_IR+CR+JRf(z) dz=0.

Getting rid of C_R
This mostly should look familiar. We parameterize C_R with z=Re^it so dz=Ri e^itdt and z²=R²e^2it. Then the integral over C_R becomes ∫_t=0^t=Π/4e^iR2e2itRi e^itdt. The modulus of this integral is bounded by ∫_t=0^t=Π/4e^Re(iR2e2it)R dt. What is the real part of iR²e^2it? We're considering i R²(cos(2t)+i sin(2t)). The real part is –R²sin(2t).
Notice that sin(2t) is increasing and concave down on [0,Π/4], and therefore sin(2t)≥4t/Π on that interval. Thus –R²sin(2t)≤–R²(4t/Π). So the modulus of the integral on C_R is overestimated by ∫₀^Π/4Re^–R2(4t/Π)dt which can be computed exactly: it is Re^–R2(4t/Π) (–R²(4/Π))^–1]₀^Π/4=(Π/{3R})(1–e^–R2). This surely →0 as R→∞.

What happens to J_R?
Here we have z={(1+i)/sqrt(2)}t, going from R to 0. So dz={(1+i)/sqrt(2)}dt and iz²=i(i)t²=–t². Therefore ∫_JRe^iz2dz=–{(1+i)/sqrt(2)}∫₀^Re^–t2dt. Now as R→∞ this integral converges to a well-known (if you don't know, I will show it to you in a second!) VALUE. So ∫₀^Re^–t2dt→ ∫₀^∞e^–t2dt=VALUE. The limit of ∫_JRe^iz2dz is –{(1+i)/sqrt(2)}VALUE as R→∞.

And the integral on I_R
Here z=t, a real parameter. Then e^it2=cos(t²)+i sin(t²), and therefore since the integral around I_R+C_R+J_R is always 0, we learn (since the C_R→0 and the J_R integral approaches a limit) that the I_R integral approaches –(–{(1+i)/sqrt(2)}VALUE). But ∫_JRe^iz2dz= ∫₀^Rcos(t²)dt+i∫₀^Rsin(t²)dt. Therefore the limits exist, and the value is Fresnel_C+iFresnel_S. so Fresnel_C and Fresnel_S are both equal to VALUE/sqrt(2).

What is this "VALUE"?
We'll be able to get this in another week of this course using complex variable methods. But here is a quick way to compute it from calc 3 which you may have seen before. We compute twice the value of VALUE:

    (∫₀^∞e^–t2dt)²=∫₀^∞e^–x2dx·∫₀^∞e^–y2dy=∫₀^∞∫₀^∞e^–x2e^–y2dx dy ∫∫_R²e^–(x2+y2)dA_x,y.
But this last double integral has circular symmetry in both the integrand and the domain so conversion to polar coordinates may be useful:
    ∫_θ=0^θ=2Π∫_r=0^r=∞e^–r2r dr dθ=2Π{–(1/2)}e^–r2]₀^∞=Π (yes, everything should have lim_r→∞). So VALUE=sqrt(Π)/2 and both Fresnel_C and Fresnel_S are Π/[2sqrt(2)].

The Fresnel integral is used in many engineering and physics applications (railroad tracks and diffraction!) and a related curve, the Cornu or Euler spiral, is essentially the only plane curve with curvature proportional to the length from a starting point (κ=s for those who remember the letters and their meaning): a curve with a simple geometric property but with a rather complicated algebraic description.

Exams were returned. Here is information about the grading of the exam.

Monday, March 1	(Lecture #12)

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Today we'll discover more about power series. We will see the reasons that power series are computationally effective. The underlying rationale is the amazing strength of GEOMETRIC SERIES. The whole subject depends on   ∑_j=0^∞ar^j=a/(1–r) for |r|<1.
The explanations I give below were probably first understood and explained by Weierstrass, an ancestor of the course instructor.

I will discuss the following statements about series of the form &sum_j=0^∞a_j(z–z₀)^j, where {a_j} is a sequence of complex constants. Such a series is said to be a power series centered at z₀. Then:

1. Each power series has a radius of convergence, R. R is one of these numbers: [0,∞]. By this I mean:
   • If R=0, then the series converges only when z=z₀. The series diverges for all other z's.
   • If R=∞, then the series converges absolutely and therefore converges for all complex numbers, z.
   • If 0<R<∞, then the series converges absolutely and therefore converges for all complex numbers, z, with |z–z₀|<R and the series diverges for all complex numbers, z, with |z–z₀|>R.
2. Suppose R≠0. Then the sum of the series, inside the radius of convergence, is a continuous function.
3. Suppose R≠0. Then the sum of the series, inside the radius of convergence, is a complex differentiable function whose derivative is the sum of the series &sum_j=0^∞j a_j(z–z₀)^j–1 which has the same radius of convergence.

We know by our experience both in calc 2 and in this course that in many specific examples, we can use methods like the Ratio and Root Tests and that the "output" from these tests is a conclusion like the first statement. And we've seen examples with R=0 and R=∞ and any R in between. What this statement declares is that any power series behaves qualitatively like the examples we've seen. The methods to get R are neat, but all the series must behave this way. There's never a square of convergence or maybe convergence in two disconnected "lumps". This simple behavior is not true for other series, but it is true for power series: easy to understand.

Here: consider the ugly and unlikely series ∑_j=1^∞(e^{(1–[Re(z)]2)})^j. This is a geometric series with a weird ratio. It converges if e^{(1–[Re(z)]2)} is less than 1. Let's see: that means the input to exp should be negative: 1–[Re(z)]²<0 or 1<[Re(z)]². So the series converges in the union of two halfplanes, {Re(z)<–1}∪{Re(z)>+1} and diverges for {–1≤Re(z)≤1}. This example is not interesting to me but it does show that the first statement does have content.

The second statement declares that the sum of a power series, inside its radius of convergence (a domain, a connected open set), is continuous. Anyone with experience with Fourier series knows that a similar statement about Fourier series is false. Fourier series can converge to terrible functions, with jumps and loathsome (!!) wiggles. Well, not loathsome, but certainly strange. This is the useful but complicated nature of Fourier series. But power series are so nice.

The third statement has a whole collection of consequences. For example, it can be applied repeatedly to show that the series can be differentiated any number of times. Then, by evaluation at z₀, the coefficients of the series can be identified with the coefficients of a Taylor series centered at z₀. Therefore any power series with R≠0 must be a Taylor series, and any function which is the sum of a power series centered at z₀ has exactly one such series, so any (legitimate!) way you get a power series for a function gives the correct answer. I'll go over this in detail.

In what follows I'll only look at the case z₀=0 because this is less writing. There will be times in this course and in applications that we can't always take z₀=0: it just won't be natural. But right now, I opt for this easier alternative.

Everything starts with the following statement:

    Convergence propagates inward
If ∑_j=0^∞a_jz^j converges for z=z₁≠0, and if
|z₂|<|z₁|, then the series converges for z₂.

This is the easiest claim, and the way to understand why it is correct is a prototype for all of the others. So we know that ∑_j=0^∞a_jz₁^j converges. Since the series converges, the terms →0. (The converse isn't true [Alternating Harmonic Series!] but the assertion this way is true.) This means that the individual terms all are a collection of numbers inside some (finite!) disc centered at 0. So there is M>0 so that |a_jz₁^j|<M. Now take some positive real number t with |z₂|<t<|z₁|. Look at the picture, please: there certainly is a lot of room to select such t. Then t=(t/|z₁|)|z₁|. The number t/|z₁| (think of it as r) is a positive number less than 1. So a geometric series with ratio t/|z₁| must converge. Therefore the series ∑_j=0^∞M(t/|z₁|)^j converges. But |a_j|t^j=|a_j|(t/|z₁|)^j|z₁|^j≤Mr^j. And further, |a_jz₂^j|<|a_jt^j|<Mr^j. The modulus of each term of the power series at z₂ is less than the corresponding term of a convergent positive geometric series. Therefore the series converges absolutely at z₂, and thus it converges: convergence propagates inward!

Showing why the first statement is true
Now let's look at the whole situation. Suppose the series converges at z₁ and diverges (doesn't converge) at w₁. Well, certainly |z₁|≤|w₁| or else we'd be contradicting the propagation of convergence. But now I know that the series converges for all z's with |z|<|z₁| and I also know that the series diverges for all w's with |w|>|w₁|. Divergence propagates outward or else we'd have a contradiction. So "color" the plane into convergent and divergent points. The only way we don't contradict the facts we now know is for these "colors" to meet up at one circle, a circle of radius R and that convergence occurs inside the circle and divergence outside the circle. If there's no such circle, we're in trouble. Notice, please, that I am not discussing the coloring of the circle itself -- I am not concerned with that here.

How the sum of the series behaves
Now let me move ahead and discuss what happens to the sum inside the radius of convergence, R. Here I will assume that R is not 0 because I want to compare values of the sum at two different points. For |z|<R, let me define f(z) to be the value of ∑_j=0^∞a_jz^j. This is, remember, the limit of the partial sums: I am still not adding up infinitely many numbers! Let me again take some t which is between 0 and |z₁| where the series converges at z₁. I will take any two complex numbers z₂ and z₃ with |z₂| and |z₃| both ≤t, and I will try to convince you that if |z₂–z₃| is small, then I can make |f(z₂)–f(z₃)| as small as I want.

Well, f(z)=∑_j=0^∞a_jz^j=∑_j=0^Na_jz^j+∑_j=N+1^∞a_jz^j. I have split the sum up into a finite sum and an "infinite tail". Remember that our z's of interest have modulus at most t. And also remember that |∑_j=N+1^∞a_jt^j|≤∑_j=N+1^∞Mr^j. This second infinite series is a geometric series, and its sum is Mr^N+1/(1–r) where r is a number less than 1. I can make this very small by selecting N large.

Showing why the second statement is true
So I want to "control" |f(z₂)–f(z₃)|. Let's say I want this to be less than some really small positive number Q. I am going to outline a method which can actually be used numerically in applications. Well, first select N so that the tail will be less than Q/3. Then, having done that, look at this description of the difference:
f(z₂)–f(z₃)=∑_j=0^∞a_jz₂^j–∑_j=0^∞a_jz₃^j=(∑_j=0^Na_jz₂^j+∑_j=N+1^∞a_jz₂^j)–(∑_j=0^Na_jz₃^j+∑_j=N+1^∞a_jz₃^j).

So I have two polynomial pieces and two infinite tails. If I drop the two infinite tails, the errors I'll make in modulus are at most 2Q/3. What about the other pieces? Well, we have ∑_j=0^Na_jz₂^j–∑_j=0^Na_jz₃^j.
These are finite sums and I'm not afraid of finite sums. In fact, this is just ∑_j=0^Na_j(z₂^j–z₃^j). Can I make this small? Now we're discussing whether polynomials are continuous. Actually, I won't go through the details (they can be found in any of our hugely thick calc 1 textbooks) but by making |z₂–z₃| very small (and also both with modulus at most t), I can make the difference of the modulus of the values of a fixed polynomial at z₂ and z₃ small. I can make the difference less than Q/3, in fact.

If you give me a specific power series, these estimates can be carried out computationally. You may not like them, but the process is quite effective.

The differentiated series
Now, finally, I want to address the third claim, about derivatives. This is more complicated. The computations can also be carried out numerically, but definitely the effort is more elaborate. Suppose that the series for f(z) converges for z₁ and, as before, t is a real positive number with t<|z₁|. So I know that ∑_j=0^∞a_jt^j converges because we compared it to ∑_j=0^∞Mr^j. But, wait: the series ∑_j=0^∞jMr^j–1 also converges absolutely! I can use the Ratio Test, for example, and since r<1 and lim_j→∞(j+1)/j=1, the extra j in front doesn't matter. Since ∑_j=0^∞jMr^j–1 converges absolutely, we know that ∑_j=0^∞j a_jt^j–1 converges absolutely by comparison. And then we know that ∑_j=0^∞j a_jz₂^j–1 converges absolutely for any z₂ with |z₂|≤t. Therefore (again, again, again!) this "termwise differentiated" series converges. Call its sum g(z). This logic, if examined closely, shows that the radius of convergence of the series ∑_j=0^∞j a_jz^j–1 is the same as that of the original series.

Are we done? No, we are not done!!!
The logic is quite involved. We have a power series ∑_j=0^∞a_jz^j which converges to f(z) when |z|<R. We have a power series ∑_j=0^∞j a_jz^j–1 which converges to g(z) when |z|<R. Why should f´(z) exist, and, if it does, why should to be equal to g(z)? This is exactly what goes wrong with things like Fourier series. You write stuff that looks good, but then if you try to verify what you really want to be true (or, better yet, compute all these things numerically) then it doesn't work.. This sort of experience which led to a great deal of confusion historically.

A detailed verification that f´ exists and equals g is in the textbook. I'll try to give some reasons to expect that this is correct. Almost every complex analysis book has slightly different reasoning here, because the sums and the related algebra are intricate. The slickest verification I know is in a book, Complex Analysis by Kunihiko Kodaira (in the math library, in the QA331 section). I don't want to present it because ... it is just too good. You read it and can't see any difficulties. So I'll be much more ordinary.

Let's start with monomials, z^j, and consider z's with |z|<t and t<|z₁| so the series for f(z₁) converges. Differentiation means that we need to compare [(z+h)^j–z^j]/h with jz^j–1 for h's with |h| very small. In fact, let me choose make h so that |h|≤SMALL=|z₁–t|/100. I don't think 100 is needed, but I don't much care -- if I were implementing this numerically, I would need to go over everything really carefully, but here I just want to give a poetic (!) impression. If you don't like algebra or binomial coefficients, go to sleep for 5 minutes because the reasoning will be very dense.

(z+h)^j=∑_k=0^jC_k^jz^j–kh^k where C_k^j are binomial coefficients. This is the simplest finite version of the Binomial Theorem which has been discovered and rediscovered in culture after culture (China, Greece, India, Persia, and, much later, France!). The beginning of the sum is interesting here:
    ∑_k=0^jC_k^jz^j–kh^k=zⁿ+nz^n–1h+h²JUNK.
If we were only interested in the differentiation of z^j (as in calc 1) we can stop here because this tells me that {[(z+h)^j–z^j]/h}–nz^j–1 is hJUNK and that as h→0 things →0. The trouble here is that we're going to be adding up linear combinations of these monomials (multipling them by the coefficients a_j) and then taking limits (we are considering infinite series!) and we need to worry about how big infinitely many JUNK terms are! This is not so easy, not only because we have infinitely many terms, but also because the binomial coefficients can get really big.

So, in more detail, {[(z+h)j–zj]/h}–nzj–1 is ∑k=2jCkjzj–khk–1. The k–1 in the exponent occurs because I'm dividing by h in the difference quotient for zj. The lower bound on the sum is k=2 because the 0 and 1 terms drop out (because of –zj, division by h, and –nzj–1). I want to estimate the modulus of this, which is |∑k=2jCkjzj–khk–1|. Of course I'll use the Δ≤ to overestimate and get ∑k=2jCkj|z|j–k|h|k–1. This is almost like the original Binomial Theorem. I really will need |h| "outside" in order to squeeze this small, so I rewrite (150 years of irritation!) the sum as |h|∑k=2jCkj|z|j–k|h|k–2. Now wait: for the |h|'s inside the sum I will do a different overestimate: |h|≤SMALL. Then the overestimated sum becomes ∑k=2jCkj|z|j–kSMALLk–2. But A–2Ak=Ak–2 so ∑k=2jCkj|z|j–kSMALLk–2=∑k=2jCkj|z|j–kSMALL–2SMALLk=SMALL–2∑k=2jCkj|z|j–kSMALLk. We can make this bigger (overestimate it) by taking the sum, which starts at k=2 and changing this lower bound to k=0. So here is something bigger: ∑k=0jCkj|z|j–kSMALLk. The Binomial Theorem tells me that I can write this more compactly as (|z|+SMALL)j. This is less than, in turn, (t+SMALL)j. Sort of showing why the third statement is true There are lots of details, which are written in your textbook. The whole thing seems like a really big cheat: we used the Binomial Theorem to unpack a power of a sum (more or less at z+h), and then we manipulate things and repack a sum into a different power (more or less at z+SMALL). Here's another thing: I can now think of t+SMALL as a new "t", let me call it t1, which is just as valid as t. And I can wrap up the whole series, just as in the very first part of this discussion, by overestimating it by |h|SMALL–2∑j=0∞|aj|(t1)j. Now the only thing "changing" is |h| as h→0, so we get a verification that the derivative of the infinite series is the infinite series of the differentiated terms because the error between the difference quotient for whole series and the infinite series for the derivative candidate is →0.

We needed the special forms of the geometric series and the Binomial Theorem. The reasoning is very specific. I will show you some common examples using Fourier series where things just don't work because these results are not available.

I needed to prepare very carefully for this lecture, unlike almost all the others in the course! This is certainly the most technical reasoning in the course, and I will not again need to refer to these estimates or these methods. But you will see why the logic of the statements we discussed today are a very important part of complex analysis.

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Wednesday, February 24	(Lecture #11)

What I did last time was remarkably subtle stuff. Experience teaching exactly that material a number of times on both the graduate and undergraduate levels has taught me that understanding what I would hope was "simple" computations instead turns out to be, for some people, rather an upsetting collection of computations and ideas. I hope you will learn from that class.

Today, I want to continue the schizophrenia (?) of the subject, and at least initially concentrate on a specific example of an analytic (indeed, entire!) funtion: sine.

What should sin(z) be? I'm not even asking whether there should be a function called sin(z), because if there is not, then we should abandon our attempts to do calculus with complex numbers. I'll outline one approach which is perhaps somewhat idiosyncratic (the noun: "a mental constitution, view or feeling, or mode of behaviour, peculiar to a person."). It turns out that there's only one answer, but many ways to "guess" the answer.

Well, sin(z)=sin(x+i y) should be sin(x)cos(i y)+cos(x)sin(i y). This doesn't seem to help too much because now I'm faced with cos(i y) and sin(i y). I could guess at thses knowledgeably just as I did earlier in the course with e^i t and learning about the differential equations they satisfy along with the initial conditions. Let me try another approach, just for fun.

What should cos(i y) be? Well, from calc 2 I know about Taylor series and about the remainder term and about ... well, lots of things. In particular, one of the simplest applications is that the infinite series 1–t²/2!+t⁴/4!–t⁶/6!+... converges for all t, and that the sum of this series is cos(t). well, we know it converges now for all complex t because we could apply, again, the Ratio Test, learn that the series converges absolutely for all complex t, and therefore must converge for all complex t. But if we insert i y for t, and take powers then we seem to get 1+y²/2!+y⁴/4!+y⁶/6!+... and maybe we recognize that these are the even powers of the exponential function. So this is {e^y+e^–y}/2. And this has a name: it is called cosh(y), the hyperbolic cosine of y. The word "cosh" has a non-technical meaning: "A weighted weapon similar to a blackjack."

Similarly, I can take sin(i y) and feed i y into the series for sine: t–t³/3!+t⁵/5!–t⁷/7!+.... The convergence discussion for this series is entirely similar to what happens with cosine, so when we plug in i y the result is i(y+y³/3!+y⁵/5!+y⁷/7!+...). What's inside these parentheses are the odd powers of the exponential function, and therefore we can get them by subtracting e^–y from e^y and dividing by 2. And don't forget the i. The sum of the series (without the i!) is called sinh(y) (hyperbolic sine, pronounced like the English word "cinch" which means "Something easy to accomplish.").

Let me steal the following table from the last time I taught Math 421, Advanced Calculus for Engineers:

What every child should know about trig and hyperbolic functions ...
TRIG onometric	cosine	cos(0)=1 cos´(0)=0	y´´=–y	cos(t)=([eit+e–it]/2)	It wiggles between –1 and 1 always
	sine	sin(0)=0 sin´(0)=1	y´´=–y	sin(t)=([eit–e–it]/2i)	It wiggles between –1 and 1 always
HYPER bolic	cosh	cosh(0)=1 cosh´(0)=0	y´´=y	cosh(t)=([et+e–t]/2)	It gets big both ways both sides are +
	sinh	sinh(0)=0 sinh´(0)=1	y´´=y	sinh(t)=([et–e–t]/2)	It gets big both ways + on the right; – on the left

Let me add further that just as the parametric curve (cos(t),sin(t)) describes the unit circle, the pair of functions (cosh(t),sinh(t)) describes a point on the curve x²–y²=1. Just plug in: (e^t+e^–t}/2)²–({e^t–e^–t}/2)²=(1/4)(e^2t+2–e^–2t–{e^2t–2–e^–2t)=4/4=1.
The picture below summarizes this situation (and no, I don't want to go into details!).

Back to (complex) sine
We know that sin(z)=sin(x)cosh(y)+i sinh(y)cos(x). We checked by direct computation that sin(x)cosh(y) and sinh(y)cos(x) are harmonic: Δ[sin(x)cosh(y)]=0 and Δ[ sinh(y)cos(x)]=0. On the way we also got evidence that the Cauchy-Riemann equations are satisfied, so that sinh(y)cos(x) is a conjugate harmonic function for sin(x)cosh(y).
So this is an analytic function, the only version of sine that people want.

The image of the real axis is the interval [–1,1] (just the usual values of sin(x) since sinh(0)=0) and the image of the imaginary axis is (x=0!) all of the imaginary axis (i sinh(y)).

When is sin(z)=0? Then the imaginary part=0 implies that either sinh(y)=0 or cos(x)=0. If sinh(y)=0, then y=0. If cos(x)=0 then x is Π multiplied by (odd integer)/2. But then sin(x) wouldn't be 0 and cosh(y) is never 0. So therefore we must have y=0 and x=2Π(an integer). Therefore sin(z) has the same roots as the usual sine function.

But could sin(z)=2? Let's see: again, sinh(y)cos(x) would have to be 0. sinh(y) could be 0 but then the real part wouldn't get big enough. So cos(x) should be 0. sin(x) will be ±1. Since cosh is always positive, let's make sin(x)=1, say with x=Π/2. Then select y so that cosh(y)=2. Can we do this? Well, we get a quadratic equation in e^y, and the result is y=ln(2±sqrt(3)). So sin[(Π/2)+i ln(2±sqrt(3))] is 2. Is this distressing? Are there other solutions? Well, yes, all separated by integer multiples of 2Π. Notice, please, that ln(2+sqrt(3)) is –ln(2–sqrt(3)), so the solutions we just found are shown to the right. Only two of a doubly infinite family of the solutions are shown!

The mapping properties of sine are new. Let me reproduce figure 1.25 of the textbook:

When x=0, we get sin(0+i y)=0+i sinh(y), so the image of the positive imaginary axis is the positive imaginary axis. For y=0 and x between 0 and Π/2, the image is the interval [0,1] on the real axis. But when x=Π/2, then we get sin(Π/2+i y)=sin(Π/2)cosh(y)+i sinh(y)cos(Π/2)=cosh(y). As y goes from 0 to ∞, the image point moves from 1 to ∞ on the positive real axis. Then some more thought shows that the image of the half strip is a quarter plane.

What happens to horizontal and vertical paths under the sine mapping? Here I didn't have time in class (but the computations are in the book (section 1.5). The horizontal line segments get changed to the first quadrant part of ellipses. As the horizontal line segments move down to the real interval [0,Π/2], the image ellipses collapse to the interval [0,1]. The vertical half lines all turn into halves of hyperbolic arcs. As we will see later due to very general theorems, since the blue and green curves generally intersect orthogonally, their images are also orthogonal. In general, angles are preserved. But a strange thing happens at Π/2. There sin(Π/2)=1 and sin´(Π/2)=cos(Π/2)=0. Locally Taylor's Theorem is still valid, and for z near Π/2, sin(z)=sin(Π/2)+sin´(Π/2)(z–Π/2)+[sin´´(Π/2)/2](z–Π/2)²+H.O.T. (that "H.O.T." is, of course, "higher order terms"). Then locally, sin(z)=1+(–1/2)(z–Π/2)²+H.O.T.. Squaring doubles angles locally, so the right angle becomes a straight angle. This is the only point in the picture where angles are not preserved under the sine mapping.

We checked by direct computation that sin(–z)=–sin(z) and that sin(z)=sin(z). Then (look at conjugation followed by –) the half strip defined by –Π/2<x<Π/2 and $y>0 gets mapped by conjugation and minusing into the second quadrant. The result is this:

Can we get the lower halfplane? Yes, since sin(–z)=–sin(z), the lower half strip gets mapped to the lower halfplane. Every vertical strip which is 2Π wide gets mapped ONTO all of C!

Other functions
Cos(z) and sinh(z) and cosh(z) are all like this. They are new and can be understood. The real problems occur with something like arcsin. We'll see if there is enough time in the course to deal with that.

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Monday, February 22	(Lecture #10)

I'll review the definitions of analytic function. A trick question: where is |z|² analytic? Since it is complex differentiable at only 1 point (the origin) there is no open set in which it is complex differentiable and therefore it is not analytic in any open set. A domain is a connected open set, and an entire function is analytic in all of C.

I'd like this statement to be correct:
    If f´(z)=0 always, then f is constant.
But it is not. Here is an example: if I have, say, a function which is defined in two disconnected blobs, and is 56 in one of them and –7i in the other. The missing ingredient is connected. So:

Theorem If f is analytic in a domain and if f´(z)=0 always, then f is constant.
A proof is that if f´(z)=0 then u_x=0 and u_y=0 and v_x=0 and v_y=0 using the Cauchy-Riemann equations. Then in a connected open set, I can connect two points by a piecewise linear path. In fact, if you look closely, I can modify this to a piecewise linear path whose pieces are parallel either to the x- or y-axes. But along either of these, u and v cannot change (apply FTC to the correct derivative -- on a horizontal sement, for example, the difference in u at the endpoints is the definite integral of u_x along that path, so the difference is 0). Then u and v cannot change along the path, so its value at a random two points must be the same.

There are simple consequences of the Cauchy-Riemann equations which are quite neat. So I can tell you that there are not very many analytic functions which are always real. In fact, if v(x,y)=0 always, then f(z) is constant. (In a domain, darn it!). Because then v_x and v_y are both always 0, and (Cauchy-Riemann equations again) we deduce that u_x and u_y are always 0, so we are reduced to the previous assertion.

Consider z², which is certainly entire. Its real part is x²–y². Is there an analytic function whose real part is just, say, x³? The answer is no. Let's see why. If u=x³, then u_x=3x² and u_y=0. Then I know (Cauchy-Riemann equations!) that v_y=3x² and v_x=0. But v_yx must be the same as v_xy, and since 6x is rarely equal to 0, this is impossible!

It is time to make a few more definitions. A real-valued function (sufficiently differentiable!) h(x,y) is called harmonic if Δh=0. By Δ I mean the following silly thing: ∂²/∂x²+∂²/∂y². This is called the Laplace operator and is very important in theory and in applications (it is the only rotationally symmetric pure second order operator).

Theorem If f(z)=u(x,y)+i v(x,y) with u and v both twice differentiable, then both u and v are harmonic functions.

Let me compute: ∂u/∂x must be ∂v/∂y (CR!). So ∂²u/∂x² is ∂²v/∂y∂x. But also ∂u/∂y is –∂v/∂x, so ∂²u/∂²y is –∂²v/∂x∂y. But mixed partials should be equal (in this courses, all of our functions will be very differentiable so we won't need to worry about bad examples here). Therefore the sum of ∂²u/∂x² and ∂²u/∂²y will be 0.
Now v is the real part of the analytic function –i f (switches u and v and puts a + sign on v) so since v is the real part of an analytic function, it also is harmonic.

Examples include x²–y² and e^xcos(y) since these are the real parts of an analytic function. But I can't just push these together and hope that the result is analytic, because (x²–y²)+i(e^xcos(y)) is not analytic. The real and imaginary parts of an analytic function must be specially linked.

Definition Suppose u and v are harmonic functions. Then v is a harmonic conjugate of u if u+i v is analytic.

So the problem I then posed students was: given a harmonic function h(x,y), how can we "create" a harmonic conjugate, g. Well, after some thought, people remarked that if the CR equations were to be true, then the hypothetical function g(x,y) would need to satisfy:
    ∂g/∂x=–∂h/∂y;    ∂g/∂y=∂h/∂x.
These are just the CR equations "backwards". Since I am supposed to know how g "changes" (g_x and g_y) I should be able to reconstruct g from its value at 1 point. For example, suppose I knew g(x₀,y₀) and wanted g(x,y). I could integrate g_x from (x₀,y₀) to (x,y₀) and then integrate g_y for (x,y₀) to (x,y). So I get the following intricate recipe (replacing the partial derivatives by the known partial derivatives of h):
    g(x,y)=g(x₀,y₀)+∫_x0^x–∂h/∂y(t,y₀) dt+∫_y0^y∂h/∂x(x,s) ds
I used t for the parameter along the horizontal line and s for the parameter along the vertical line. I would like to verify that the function g I created has the correct partial derivatives. Well, the formula I just wrote only has y appearing once and there as an upper parameter of a definite integral. Therefore ∂g/∂y is ∂h/∂x(x,y) using FTC on the second integral (the first integral has no y in it so its derivative is 0 with respect to y). I would like to check the x derivative, but maybe I could just write a different integral, like this:
    g(x,y)=g(x₀,y₀)+∫_y0^y∂h/∂x(x₀,s) ds+∫_x0^x–∂h/∂y(t,y) ds
Now, using this formula, there is only one appearance of x, and I can compute using FTC again to get ∂g/∂x=–∂h/∂y.

So I verified that the partial derivatives are what they should be and I have "created" a harmonic conjugate. The only difficulty is that I used two different formulas for the same quantity, g(x,y), and why should these formulas give the same result?

They give the same result if the integral around the two paths is the same. Wait: this is the same thing as asking that the integral all the way around the (correctly oriented!) edge of the rectangle is 0. Let me subtract off the g(x₀,y₀) from both of the formulas, and reverse the orientation of the top & left path. Then I want to compute the line integral:
    ∫_C(–∂h/∂y)dx+(∂h/∂x)dy
where C is the boundary indicated of the rectangle. Maybe we should use Green's Theorem with P(x,y)=–∂h/∂y and Q(x,y)=∂h/∂x. The integrand in the double integral side of Green's Theorem is ∂Q/∂x–∂P/∂y.
    If Q(x,y)=∂h/∂x, then ∂Q/∂x=∂²h/∂²x.
    If P(x,y)=–∂h/∂y, then –∂P/∂y=–(–∂²h/∂²y).
The sum of the right-hand sides is Δh which is 0 since h is harmonic so the double integral is actually integrating 0, and the line integral side of Green's Theorem is 0 also. Therefore the two integrals are computing the same thing, and we are correct.

A specific example
So I tried to consider h(x,y)=ln(x²+y²). I first claimed this was harmonic and verified it:
    h_x=–2x/(x²+y²).
    h_y=–2y/(x²+y²).
Now the second derivatives:
    h_xx=[–2(x²+y²)–2x(–2x)]/(x²+y²)²=(2x²–2y²)/(x²+y²)².
    h_yy=[–2(x²+y²)–2y(–2y)]/(x²+y²)²=(2y²–2x²)/(x²+y²)².
It is almost amazing that Δh=∂²h/∂x²+∂²h/∂x²=0 since the tops cancel.

How to "create" (attempt to create) a harmonic conjugate
I switch x and y derivatives and insert a minus sign. The line integral I should consider is (–∂h/∂y)dx+(∂h/∂x)dy which is 2y/(x²+y²)dx–2x/(x²+y²)dy. I will assume for simplicity that the harmonic conjugate I am attempting to create for this h(x,y) has value equal to 0 at (–1,–1), and I will try to compute (actually compute, actually get a value!) for h(1,1). Well, if you consider the accompanying diagram, one way I could get from (–1,–1) to (1,1) is by way of the piecewise linear path ABC. So this integral is
     ∫_x=–1^x=1–2/(x²+1)dx+∫_y=–1^y=1–2/(1+y²)dy.
This is because y=–1 on AB and dy=0, and, on BC, x=1 and dx=0. I can compute these integrals. They are both arctan integrals, and the value is –2Π.

I also could get from (–1,–1) to (1,1) is by way of the piecewise linear path ADC. So this integral is
     ∫_y=–1^y=12/(1+y²)dy+∫_x=–1^x=12/(x²+1)dx.
This is because x=–1 on AD and dx=0, and, on DC, y=1 and dy=0. I can compute these integrals. They are both arctan integrals, and the value is 2Π.

These are supposed to be equal! What is the problem?

People thought about this, and several people volunteered that Green's Theorem could not be applied with the specific second example I was trying since ln(x²+y²) was not even defined at (0,0) which is inside the square. Of course, that's the catch -- it is easy to be deceived by formulas. I wrote precisely what I had actually done:

TheoremSuppose h(x,y) is a harmonic function defined in all of a disc (no holes!!!). Then h(x,y) has a harmonic conjugate g(x,y) defined in that disc, g(x,y) is harmonic in that disc, and h(x,y)Ii g(x,y) is analytic in all of that disc.

Example The function ln(x²+y²) is harmonic in all of C except for 0. On that domain (it is a connected open set, even though it has a "hole") the function ln(x²+y²) has no harmonic conjugate.
This function is the real part of (1/2)log(z).

These are not obvious computations and not obvious ideas. They took a great deal of time to develop.

As a final example, I pulled the function x³–3xy² out of a hat (must have been a remarkable hat!). I verified that it was harmonic, and then just guessed (hah!) at a harmonic conjugate: 3x²y–y³. Of course, I didn't guess. These functions are the real and imaginary parts of the entire function z³.

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Friday, February 19	(Lecture #9)

SNOW DAY MAKE UP!!!
We stated this result last time.
Theorem Suppose u(x,y) and v(x,y) have continuous first partial derivatives. Then lim_Δz→0(u(x+Δx,y+Δy)+i v(x+Δx,y+Δy))–(u(x,y)+i v(x,y))/Δz exists exactly when (o.k., "if and only if") u_x=v_y and u_y=–v_x.

We showed that if f is complex differentiable, then the Cauchy-Riemann equations must be true. We actually saw that the derivative, f´(z), is equal to both (∂u/∂x)+i(∂v/∂x) and (∂v/∂y) i(∂u/∂x).

Why is the converse assertion in the theorem true (reversing the logic if...then)? Remember the following which is stated in calc 3 but rarely verified, even though it only needs the Mean Value Theorem of calc 1. If Q(x,y) has first partial derivatives which are continuous, then Q(x+Δx,y+Δy)=Q(x,y)+{∂Q/∂x}Δx+{∂Q/∂y}Δy+H.O.T.. The H.O.T. stands for "higher order terms" and stands for "stuff" that →0 faster than any first order term in Δx and Δy. For example, lim_Δx→0H.O.T./Δx=0 and lim_{sqrt((Δx)²+(Δy)²)→0} H.O.T./sqrt((Δx)²+(Δy)²)=0. Now I'll use this applied to both the real and imaginary parts of f.

f(z+Δz)=u(x+Δx,y+Δy)+i v(x+Δx,y+Δy)=
    u(x,y)+{∂u/∂x}Δx+{∂u/∂y}Δy+H.O.T.₁+i(v(x,y)+{∂v/∂x}Δx+{∂v/∂y}Δy+H.O.T.₂)
This is very complicated but let me rethink a bit. First, I will combine the two H.O.T.'s because the distinction between them won't matter. Second, if I am assuming the Cauchy-Riemann equations, then I could call {∂u/∂x}={∂v/∂y} by one name, let's say A, and I could call {∂u/∂y}=–{∂v/∂x} by another name, let's say –B (so {∂v/∂x} is B, of course!). Then the material above becomes this:
    u(x,y)+AΔx–BΔy+i(v(x,y)+BΔx+AΔy)+H.O.T.
But let me recognize a bit more, if I realize that (A+iB)Δz=(A+iB)(Δx+iΔy)=(AΔx–BΔy)+ (AΔy+BΔx) then the preceding becomes
    u(x,y)+i v(x,y)+(A+i B)(Δx+iΔy)+H.O.T.
Finally, rewrite in terms of z's and f's:
    f(z+Δz)=f(z)+(A+i B)Δz+H.O.T.
This means that the quotient {f(z+Δz)–f(z)}/Δz is A+i B+{H.O.T./Δz}. As Δz→0, the last term also →0 because the top is "higher order". So the limit exists and is A+i B. We have verified the claim above: only the Cauchy-Riemann equations are needed to insure complex differentiability.

Since the definition of complex differentiability is exactly like real differentiability, I bet that the sum, product, quotient, composition, etc. of complex differentiable functions is complex differentiable, and that the derivatives obey the same algorithms as in "real" calculus.

Complex differentiability turns out not to be exactly the correct concept. For example, we saw that |z|² is complex differentiable, but only at z=0. That turns out to be a peculiarity, and we need a bit more information. The key definition is the following:

Key definition
Suppose U is a connected open set (a domain). Then a function f defined on U is said to be complex analytic or holomorphic if f is complex differentiable at all points of U. We will also assume (this turns out to be logically not needed!) that f´ is continuous. So a holomorphic function is one which is defined and has a continuous complex derivative in a connected open set. These are the functions discussed in this course.

I want to show you that already I know lots of things, so I will compute infinitely many line integrals. First, though, some examples.

∫_|z|=1x dz
Sometimes people omit the direction of a closed curve over which a line integral is to be computed. In that case, almost always what's meant is the counterclockwise orientation, so the "inside" is a bounded region. What can I do here? I know that dz=dx+i dy, so we need to compute ∫_|z|=1x dx+i x dy. This is ideal for applying Green's Theorem with P=x and Q=i x. Then ∂Q/∂x=i and ∂P/∂y=0, so we need to compute the double integral of i over the inside of the unit circle. This is i multiplied by the area, so it is Πi. You can also, if you wish, compute this with a parameterization, etc.

∫_|z|=1(1/z) dz
Maybe this is more interesting. Perhaps I would like to apply Green's Theorem again. BUT 1/z is not continuous everywhere inside the contour of integration. Hey: it isn't defined at z=0! For this computation, I will go low-tech, and just parameterize. The unit circle can be described by z=e^i t with t in [0,2Π] and dz=i e^i tdt. The integrand, 1/z, is e ^i t so that ∫_|z|=1(1/z) dz becomes ∫_t=0^t=2Πi e^i te ^i tdt, or just ∫₀^2Πi dt=2Π i.

∫_Something!e^{(e17z5(58z5 9z3+2))} dz
Well, let's try to make Something! a more challenging curve. So: it will be a closed piecewise-linear contour, from 1+i to 2i to 2i 2 to  2 i to  3i to  1+i to 1+i. This is 6 line segments, and a moderately bizarre function. Still, I can tell you the value of this integral, and many, many others: the value is 0.

Of course I am concealing one of the most important results of the course through excessive detail: this should be a major intellectual crime. In fact, the setting has a name.

Cauchy's Theorem
Suppose f is holomorphic in U, and C is a closed curve in U whose interior is entirely in U. Then ∫_Cf(z) dz=0.

Proof This is a major result in mathematics, and a large number of people have spent a large amount of time over the last 150 years making a verification of this result understandable. So here goes.
∫_Cf(z) dz= ∫_C(u(x,y)+i v(x,y))(dx+i dy)= ∫_C(u(x,y)+i v(x,y))dx+(–v(x,y)+i u(x,y))dy.
What I used was (i)²=–1 and I reorganized the product so that I have P(x,y)dx+Q(x,y)dy in preparation for using Green's Theorem. I need to compute certain partial derivatives in order to get the integrand for the double integral side of Green's Theorem.
    Since P(x,y)=u(x,y)+i v(x,y), ∂P/∂y=∂u/∂y+i ∂v/∂y.
    Since Q(x,y)=–v(x,y)+i u(x,y), ∂Q/∂x=–∂v/∂x+i ∂u/∂x.
The integrand in the double integral is ∂Q/∂x–∂P/∂y. This is (–∂v/∂x+i ∂u/∂x ) – (∂u/∂y+i ∂v/∂y ).
We can rearrange this to get (–∂v/∂x–∂u/∂y) +i (∂u/∂x–∂v/∂y ).
Each of these real and imaginary parts is 0. These conditions are exactly the Cauchy-Riemann equations.
After 150 years, the proof should be easy, shouldn't it?

z^z
I'll try to do this problem, remembering that our definition of A^B is e^B log(A) (that's the small l version of log and the distinction is important!). The useful diagram to the left is the unit circle in the first quadrant. Of course, A is 1, B is (1+i)/sqrt(2), and C is i. I'll define S(z)=z^z -- S for "superexponential".

a) What are all values of S(A), S(B), and S(C)?
The detailed results are perhaps slightly surprising. For all of A, B, and C, ln(|z|)=ln(1)=1 which will simplify things a bit.
S(A)=e^1 log(1)=e^0+i arg(1). Now arg of any non-zero complex number has infinitely many values, so we almost expect confusion. But here arg(1)=2Πk where k is any integer, positive or negative or 0, and exp is 2Πi periodic, e⁰=1, so 1¹ has only one value, 1.
S(B)=[(1+i)/sqrt(2)]^{(1+i)/sqrt(2)}=e^{(1+i)/sqrt(2)[ln(1)+i arg((1+i)/sqrt(2))]}=e^{(1+i)/sqrt(2)[ln(1)+i((Π/4)+2Πk)]}=e^{–((Π/4sqrt(2))+2Πk/sqrt(2))}e^{i(((Π/4)+2Πk)/sqrt(2))}. I write things this way because I hope to understand e^Ke^i L if K and L are real numbers. In particular, I can easily write the real and imaginary parts of the complex number, since e^i L is cos(L)+i sin(L). So the values of S(B) are (e^{–((Π/4sqrt(2))+2Πk/sqrt(2))}cos1(((Π/4)+2Πk)/sqrt(2)) )+i (e^{–((Π/4sqrt(2))+2Πk/sqrt(2))}sin(((Π/4)+2Πk)/sqrt(2)) ), a complicated mess.
The values of S(C) were a bit surprising to me: iⁱ=e^i log(i)=e^{i[ln(|i|)+i arg(i)]} =e^{–Π/2–2Πk}. All of the values are real.

b) If we start with k=0 at 1 and move along the curve so that z^z varies continuously -- this depends on the value of arg selected. I would have been happy if students had mentioned that Arg (capital A!) was continuous on the curve (in fact, in the right half-plane, it is arctan(y/x), certainly continuous. So the values of S(B) and S(C) are the values above with k=0. Please note that if the circular arc were continued around to, say, –i, then k=0 would not be correct! The value would not be the "principle" value, but the value used when k=1. This is complicated and either irritating or interesting.

c) What does the image of the curve under the mapping z→S(z) look like? This is what I did:
S(z)=S(e^it) for z on the curve, with 0≤t≤Π/2. Then (e^it)^eit=e^{eit[log(eit)]}=e^eiti t because Arg(e^it) is just t for those points on the curve and ln(|e^it|)=ln(1)=0. Then e^eiti t=e^{i t[cos(t)+i sin(t)]}=e^–t sin(t)e^i t cos(t)=e^–t sin(t)cos(t cos(t))+ie^–t sin(t)sin(t cos(t)). I finally gave up and asked Maple which produced the graph to the right. Superior people could probably have drawn the graph with pure thought.
Please note that the "top" of this curve does not correspond to S(B). The intercepts on the real axis are at 1 and e^–Π/2, the principal values of S(A) and S(C).

The frog ...
So here is the problem statement:

A frog starts at the origin. It leaps one unit eastward (to the right!) on its first jump, 1/2 unit on its second, 1/4 unit on its third, 1/8 on the fourth, and so on, each time turning exactly an angle θ to the left from the previous jump. Assuming only that 0<θ<Π, show that this frog will always end up at some point on a semicircle of radius 2/3. Sketch this semicircle.

So the hops of the frog are the sum of a series:
    1+e^iθ/2+e^2iθ/4+e^3iθ/8+...
The e^iθ's rotate by the angle needed, and the divisions by powers of 2 shrink the lengths of the leaps. Then this is a geometric series with first term=1 and ratio between successive terms equal to e^iθ. The modulus of the ratio is 1/2 which is <1, so the series converges and its sum is 1/(1–e^iθ/2).

I followed the solution of Ms. Addabbo in what I told the class. My own solution was quite clumsy. So: she set θ=0 and θ=Π to get points on the positive real axis: 2 and 2/3. Then she guessed (?) that the semicircle would be centered at 4/3 with radius 2/3. So all she needed to do was to show that 1/(1–e^iθ/2) has distance 2/3 from 4/3. So:

   1       4      2       4    6–4(2–eiθ)
------- – --- = ------ – --- = ----------
1–eiθ/2    3     2–eiθ    3      6–3eiθ

The top becomes 2+4e^iθ. If we want the modulus of this to be 2/3, we could multiply the fraction by 3/2 and verify that it has modulus 1. So: The top becomes 6+12e^iθ while the bottom becomes 12–6e^iθ. Let's throw out the factors of 6 to get 1+2e^iθ divided by 2–e^iθ. But the modulus of the top is the same as its product with e^–iθ. That is e^–iθ+2. The complex conjugate of that is e^iθ+2, the bottom. So the modulus of the quotient is equal to 1 (the quotient of w divided by w has modulus 1 because modulus of a quotient is the quotient of the moduli and the modulus of the conjugate is the same as the modulus of the original complex number).
This is certainly not clear to me. I will show you a much more systematic way to analyze such things later in the course.

---

Wednesday, February 17	(Lecture #8)

Another way to compute some integrals
For example, suppose we have a curve, C, with z=x(t)+i y(t), for t in [a,b]. Then dz=(x´(t)+i y´(t))dt, either by differentiating that equation or by using dz=dx+i dy. If we want to integrate z⁵ on this curve, we could substitute and get ∫_t=a^t=b(x(t)+i y(t))⁵ (x´(t)+i y´(t))dt. But I know the Chain Rule: the derivative of (1/6)(x(t)+i y(t))⁶ with respect to t is exactly (x(t)+i y(t))⁵ (x´(t)+i y´(t)). So I can antidifferentiate and plug in. What I get is (1/6)(z at the end of C where t=b)⁶–(1/6)(z at the start of C where t=a)⁶.

You may recognize this as a variation of one of the calc 3 methods for computing a line integral (exact differentials and path independence). But in practice it is easy: take any curve W joining, say, 1–i and 2+3i. Then ∫_Wz⁵dz=(1/6)(2+3i)⁶–(1/6)(1–i)⁶. Very easy!

Estimating the size of complex line integrals
Many of our exact computations will depend on what seems like sloppy overestimates of the modulus of complex line integrals. Let me show you how this works, and then do a few examples. We'll have many more examples later in the course! The explanation is a bit devious and complicated, but the final result will be easy to use. I promise.

So I want to estimate I=∫_Cf(z) dz where C is a piecewise smooth curve and f(z) is continuous on the curve C. Well, I is a complex number. If it isn't 0 (in which case estimating it is rather easy) then there is θ so that e^i θ is real and positive. So now we know:
    |∫_Cf(z) dz|=e^i θ∫_Cf(z) dz=∫_Ce^i θf(z) dz=∫_t=a^t=be^i θf(z)(x´(t)+i y´(t))dt
Now, wait! The quantities above are all real and positive. But "Re" is linear and so is "∫". What I mean is the following: FROG and TOAD are real functions, then Re(∫FROG+iTOAD) is Re(∫FROG+i∫TOAD) which is Re(∫FROG). This is the same as ∫Re(FROG+iTOAD).

Since ∫_t=a^t=be^i θf(z)(x´(t)+i y´(t))dt is real and positive, it is equal to its real part: Re(∫_t=a^t=be^i θf(z)(x´(t)+i y´(t))dt)=∫_t=a^t=bRe(e^i θf(z)(x´(t)+i y´(t)))dt. But remember that the real part of anything is ≤ the modulus, so this is ≤∫_t=a^t=b|e^i θf(z)(x´(t)+i y´(t))|dt.

What are we integrating?
    |e^i θf(z)(x´(t)+i y´(t))| =| e^i θ| |f(z)| |x´(t)+i y´(t)|=|f(z)| sqrt(x´(t)²+ y´(t)²)
Now suppose some wonderful person tells me that |f(z)| is at most M on the curve. Then the quantity above is at most M sqrt(x´(t)²+ y´(t)²). Don't panic -- we're almost done. So pasting together the very start with this we get:
    |∫_Cf(z) dz|≤∫_t=a^t=bM sqrt(x´(t)²+ y´(t)²)dt.
But the integral of that strange square root quantity is exactly the length of the curve C. We have verified (and won't ever think again about the tricky verification of):

The ML inequality Suppose C is a piecewise smooth curve whose length is L, and suppose that f(z) is a function which is continuous on C with modulus bounded by M on C. Then |∫_Cf(z) dz|≤ML.

A few examples follow. These examples may seem strange, but they are motivated by many aspects of complex analysis, as you will see.

Example #1
Suppose f(z) is z²/(z+4) and C is the left half of a circle of radius R centered at  0 where R is very large. C is oriented counterclockwise, of course, the way it should be! Then |f(z)|≤|z|²/(|z|–4). I choose to underestimate |z–4| by |z|–4 rather than 4–|z| because I was told that R=|z| is "large" and this way the underestimate is positive. This is overestimated by M=R²/(R–4). The length of this curve is, of course, L=ΠR. So my estimate is ΠR³/(R–4).

Comment My pal Maple claims to compute this integral exactly (I haven't done it!). I used the Re^i t parameterization and got a definite integral. The incredibly silly exact value claimed is 8R+16Πi+16 ln({R–4}/{R+4}). So this is O(R) (big-oh of R, the order of increase as R→∞). Our estimate is much "weaker": it is O(R²).

Example #2
Here let's consider f(z)=e^i z/z³ where C is the line segment from the large positive real number R to the number R i. The length of this line segment is L=sqrt(2)R. What about M? Well, |f(z)|=|e^i z|/|z|³. The bottom is easy enough: it is R³. What can I say about |e^i z|? Realize, please, that if w=c+d i is a complex number with real and imaginary parts c and d, respectively, then e^w=e^ce^i d. The number e^i d is on the unit circle and so has modulus 1. And therefore |e^w|=e^Re(w). Here we have e^i(x+i y), so the modulus is e^–y. If we realize that C is in the upper half-plane, then we know that e^–y≤e⁰=1. Therefore a valid M is 1/R³, and our ML estimate is just sqrt(2)/R². Immediately we see that the limit of ∫_Cf(z) dz as R→∞ is 0. That's the sort of behavior we'll want.

Comment How about "exact computation"? My pal Maple does not like this integral too much. It claims, after some "thought", that its "value" is an expression with 14 terms, several of which are some quite complicated "special functions". Yes, they are hypergeometric functions, but more elaborate than what I showed you. For example, one piece involves values of ₂F₂(1,1;2,2;z)! Knowing this "exact value" doesn't help me, at least, to understand that the limit as R gets large will be 0. But such asymptotic information is important in this subject.

We are now ready to start the course.

So we need to look at {f(z+Δz)–f(z)}/Δz and of course consider what happens when Δz→0.

GOOD
Take f(z)=z². Then, just as in times past (!) we see that {f(z+Δz)–f(z)}/Δz is {(z+Δz)²–z²}/Δz which becomes 2z+Δz. When Δz→0, this becomes 2z. Indeed, the limit exists and is always 2z.

BAD
Let's try f(z)=z. I will work with some cleverness. What happens if Δz is just Δx, only a real "increment"? Well, (write z=x+i i) f(z+Δz)=f((x+i y)+Δx)=x–i y+Δx, so {f(z+Δz)–f(z)}=x–i y+Δx–(x–i y)=Δx. Then the limit of the difference quotient will be 1.

Now let's try a "purely imaginery" Δz, so Δz is just i Δy. What happens? If z=x+i y, f(z+Δz)=f((x+i y)+i Δy)=x–i y–i Δy, and the difference quotient {f(z+Δz)–f(z)}/Δz becomes {(x–i y–i Δy)–(x–i y)}/(iΔy) becomes –1.

But this means that the limit of {f(z+Δz)–f(z)}/Δz does not exist. This function is not complex differentiable!

What's going on? Apparantly the most modest requirement, that the "obvious" limit of the difference quotient exists, throws out at least one (and in fact, many!) nice-looking functions. Maybe a bit of generality is useful here, a sort of overview.

Suppose we have a function which takes the complex number x+i y to the complex number u(x,y)+i v(x,y). I want the limit of

(u(x+Δx,y+Δy)+i v(x+Δx,y+Δy))–(u(x,y)+i v(x,y)) 
-----------------------------------------------
                    Δx+iΔy

as Δx+iΔy→0 to exist. Let me temporarily call the limit A+Bi. There are some obvious special cases to consider when thinking about this limit similar to what I did during the second example above.

Set Δy=0 (a sort of horizontal limit). Then if the limit exists (as Δx→0), it must be (∂u/∂x)+i (∂v/∂x) because what's left are just two limits with Δx's.

Set Δx=0 (a vertical limit). Then if the limit exists (as Δy→0), it must be (a bit of care because there is another i in the bottom!) (1/i)(∂u/∂y)+(∂v/∂y) because what's left are just two limits with Δy's.

Now both of these are supposed to be the same complex number, A+Bi. Two ways of writing A tell me that:
    ∂u/∂x=∂v/∂y
And then two ways of writing B tell me that (remember, 1/i is –i)
    ∂v/∂x=–∂u/∂y

It is natural to wonder if there are any further "compatibility conditions" that u and v need to satisfy. For example, we could consider the limit where Δz=Δt+i Δt as Δt→0 (a sort of diagonal limit) or Δt+i (Δt)2 as Δt→0 (a sort of limit along a parabolic path). There are results for both of these but the results turn out to be linear combinations of the two already obtained. In fact, for reasonable functions, there aren't any further conditions.

Theorem Suppose u(x,y) and v(x,y) have continuous first partial derivatives. Then lim_Δz→0(u(x+Δx,y+Δy)+i v(x+Δx,y+Δy))–(u(x,y)+i v(x,y))/Δz exists exactly when (o.k., "if and only if") u_x=v_y and u_y=–v_x.

The partial differential equations u_x=v_y and u_y=–v_x are called the Cauchy-Riemann equations.

Example
Let's try a fairly benign-looking function, |z|²=x²+y². In this case, u(x,y)=x²+y² and v(x,y)=0 (the zero function whose value is 0 for all inputs). Then u_x=v_y and u_y=–v_x become 2x=0 and 2y=0 so this function is "complex differentiable" only when z=0!

Example
We tried u(x,y)=x² and v(x,y)=2xy. Then (perhaps remarkably!) the Cauchy-Riemann equations are always true, and (x²–y²)+i(2xy) is always complex differentiable. But maybe this isn't remarkable, since these functions are the real and imaginary parts of z².
Example
Now u(x,y)=e^xcos(y) and v(x,y)=e^xsin(y). Here the minus signs are neat, and again it turns out that the Cauchy-Riemann equations are satisfied everywhere. Of course, this function turns out to be e^z.

• Infinite series (especially power series)
• Line integrals

Phrase	Proper definition	Possible picture
The last two definitions (continuing from the previous lecture)
Adding curves	Suppose C1(t) is a curve with domain [a,b] and C2(t) is a curve with domain [c,d]. If C1(b)=C2(c) (that is, C1's "Head"=C2's "Tail", perhaps) then the curve C(t) whose domain is [a,b+d–c] and defined by C(t)=C1(t) if t<b and C(t)=C2(t–b+c) (I hope!) is called the sum of C1 and C2. We will write C=C1+C2.
Reversing curves	If C(t) is a curve with domain [a,b], then the "reversal of C(t)", also called –C, will be the curve, let me temporarily call it D(t), whose domain will also be [a,b], defined by D(t)=C(b+a–t). –C is just C backwards. The picture is slightly weird (how should I have drawn it?) but I hope you get the idea.

I hope the darn formulas for C₁+C₂ and –C are correct. I don't think I will ever be so formal again in this course. I will usually be rather casual. The curves in this course will all be piecewise differentiable, and, actually, I think every curve I'll consider will turn out to be sums of line segments and circular arcs. So bah to general theory!

The truth: parameterized curves
Actually I should acknowledge that what we are defining are called parameterized curves. For example, the curves C₁(t)=3t+5t i for t in [0,4] and C₂(s)=3s²+5s² i for s in [0,2] are really different, but they "look" the same: the line segment from 0 to 12+20i. But we will mostly be interested in line integrals and it will turn out that the values of such integrals will not depend upon parameterizations.

A few examples

This curve is a straight line segment starting at –4i and going to –2i, followed by a quarter circle whose center seems to be 0, starting at –2i and going to 2. If I wanted to be totally strict, I guess I could define a C(t) on the interval from [0,2+(Π/2)] Let's see: for t in [0,2), let C(t) be (t–4)i, and for i in [2,2+(Π/2)], define C(t) to be 2ei(t–2–(Π/2)). Some remarks Almost always in this course I will parameterize circles and circular arcs using the complex exponential. My first guess will always be CENTER+(RADIUS)e(i parameter). Another remark: what I've just done is excessively precise and maybe slightly silly. I want to write curves in a way that is convenient. So this curve should be written as a sum of two curves, say C1+C2. Here C1(t)=ti where t is in the interval [–4Π,–2Π], and C2(t)=0+2eit where t is in the interval [–Π/2,0]. It is just easier to do this than to try to figure out the total length, how to combine and adjust parameterizations, etc.

Here I will write the curve again as a sum, C1+C2. C1 seems to be a semicircle of radius 2 and center –1. It goes backwards. So C1(t)=–1+2ei([Π]–t) where t is in the interval [0,Π]. C2 is a straight line segment. This I know how to parameterize from thinking about convexity. C2(t)=(1–t)(1)+(t)(–3i) where t is in the interval [0,1]. A remark Parameterizing a circular arc in a clockwise fashion feels close to obscene to me. Almost always in this course our circles will be parameterized counterclockwise.

The definition
Now I want to define ∫_CP(x,y)dx+Q(x,y)dy where P(x,y) and Q(x,y) are at least continuous in an open set containing the curve C. In fact, in this course, these functions will almost always be differentiable. They will also almost always be rather explicit, as will the curve C. So here is the official definition:
If C is the curve defined by x(t) and y(t) for t in [a,b], then ∫_CP(x,y) dx+Q(x,y) dy is defined to be ∫_t=a^t=bP(x(t),y(t))x´(t)+Q(x(t),y(t))y´(t) dt.
This is of course valid if the curve is smooth. When the curve is piecewise smooth, we break up the interval [a,b] into subintervals where the functions are smooth, compute using the definition above in each of the subintervals, and then add the result.

Now there's a serious difficulty, because line integrals were originally developed to compute physical quantities which seemed to only depend on the geometry of the situation and the values of P and Q (the simplest examples are work and flux). But the definition we wrote has a dependence on the parameter, which makes no sense physically. Let's look at a specific case more closely. For simplicity I will write things only with real functions. "Complexifying" the line integrals will cause no difficulty and we'll do it later.

We connect the point (0,0) and (12,20) with a straight line segment. This can be parameterized in many ways. Let me look at the two we previously mentioned.

• Here the C₁ parameterization is given by x(t)=3t and y(t)=5t for t on [0,4]. Then &int_C1P(x,y) dx+Q(x,y) dy becomes ∫₀⁴{P(3t,5t)3+Q(3t,5t)5}dt.
• Here the C₂ parameterization is given by x(s)=3s² and y(s)=5s for s on [0,2]. Then &int_C2P(x,y) dx+Q(x,y) dy becomes ∫₀²{P(3s²,5s²)6s+Q(3s²,5s²)10s}ds.

    The value of a line integral does not depend on parameterization.

Further properties of line integrals
The following "linearity" results are easy, and have been covered in calc 3:     Additivity ∫_C1+C2P(x,y) dx+Q(x,y) dy=∫_C1P(x,y) dx+Q(x,y) dy+∫C₂P(x,y) dx+Q(x,y) dy     Reversal ∫_–CP(x,y) dx+Q(x,y) dy=–∫_CP(x,y) dx+Q(x,y) dy

Green's Theorem
The major result in the subject is Green's Theorem which connects line integrals to double integrals. The hypotheses are rather important in both applications (electromagnetism) and theory (complex variables). Here we go:

Green's Theorem Suppose C is a simple closed piecewise smooth curve which is the boundary of a connected open set R. C is parameterized so that the increasing values of the parameter have R to the left. Suppose P(x,y) and Q(x,y) are continuously differentiable functions of two variables at all points of both R and C. Then ∫_CP(x,y) dx+Q(x,y) dy=∫∫_R{∂Q/∂x}–{∂P/∂y} dA_x,y.

Some of the phrases in this result will occur again and again in this course. There are abbreviations for them. For example, a "connected open set" will be called a domain. And "C is parameterized so that the increasing values of the parameter have R to the left" will be called the boundary of R is oriented positively (actually, we probably won't even mention this most of the time, because essentially every boundary curve will be oriented positively).

People have made careers out of proving various versions of Green's Theorem with slightly weakened hypotheses (some of these have been moticated by authentic applications in physics, though). A realistic version of Green's Theorem, as quoted above, would probably take 3 or 4 lectures to prove but I'd like to verify some specific version because this gives an idea of what's going on.

I'll specify the boundary of the domain first. I want it to be roughly triangular in shape, but let me make one part of the boundary curved so some small difficulty will occur. Here we go: the boundary will be a straight line segment connecting (0,0) to (1,0), and then a straight line segment connecting (1,0) to (1,3), and then from (1,3) to (0,0), the curve y=3x⁴. The domain R is what's inside. So I want to connect the double integral to the line integrals. Let me consider first the line integrals over the boundary. There are three chunks of boundary, and I will analyze each of them.

1. C₁, the straight line segment connecting (0,0) and (1,0).
   So here I will use v=x as my parameter. v varies from 0 to 1. Notice that y=0 always here, so that ∫_C1P(x,y) dx+Q(x,y) dy becomes ∫₀¹P(v,0) dv (the Q term drops out because the derivative of 0 is 0).
2. C₂, the straight line segment connecting (1,0) and (1,3).
   Here use q=y as the parameter, going from 0 to 3. Of course x=1 always, so its derivative will be 0 and now the P term goes away. Now ∫_C2P(x,y) dx+Q(x,y) dy is ∫₀³Q(1,q) dq.
3. C₃, the curve y=3x⁴ oriented from (1,3) to (0,0).
   I'll compute it from (0,0) to (1,3) and then multiply by –1. Since y=3x⁴, I can parameterize by x=w and y=3w⁴ with w going from 0 to 1. Then ∫_–C3P(x,y) dx+Q(x,y) dy is ∫₀¹P(w,3w⁴)+Q(w,3w⁴)12w³ dw and what we need is minus that.

No accident, this matches the P part of the line integral side of Green's Theorem.

If we start with the Q part of the double integral (I'll go the other way just for fun), then
    ∫∫_R{∂Q/∂x} dA_x,y=∫_y=0^y=3∫_x=y^1/4/3^x=1{∂Q/∂x} dx dy=∫_y=0^y=3Q(x,y)]_x=y^1/4/3^x=1 dy=∫_y=0^y=3Q(1,y)–Q(y^1/4/3,y) dy.
So we have ∫_y=0^y=3Q(1,y)–Q(y^1/4/3,y) dy and should match that somehow with the line integral contribution, ∫₀³Q(1,q) dq–∫₀¹Q(w,3w⁴)12w³ dw. The first part of each expression has equal values. But why should ∫_y=0^y=3Q(y^1/4/3,y) dy and ∫₀¹Q(w,3w⁴)12w³ dw be equal? Make the substitution y=3w⁴ in the second integral and follow through the limits and dy=12w³dw etc. They are the same!

We have proved Green's Theorem for this region and its boundary. I hope you see where we needed to know about P and Q on the boundary and at every point inside. Just omitting the hypotheses at one interior point has profound consequences both practical (electric charges!) and theoretical (we'll see this!).

About George Green
George Green lived from 1793 to 1841. Biographical information is available here and here. Some aspects of his adult life were quite unconventional, but perhaps more interesting is that (quoted from the first link), "The younger Green only had about one year of formal schooling as a child, between the ages of 8 and 9." Later, in 1832, nearly 40 years old (!), he "went to college" (Cambridge). A version of his original 1828 publication, An Essay on the Application of mathematical Analysis to the theories of Electricity and Magnetism. (83 pages), is available here if you have the curiosity and courage to look at it. Almost every page is dominated by physical discussions. A modern advanced point of view is in the book Calculus on Manifolds by Michael Spivak (not an easy book, almost formidably abstract and mathematical, perhaps a counterbalance to Green's original essay).

Complex line integrals
Everything works. What do I mean? Let me take one of our previous examples, shown again to the right, and call it C. I want to actually compute the integral ∫_C3x²–i y dz. The only abbreviation I need to tell you is that dz means dx+i dy. So we want to compute ∫_C(3x²–i y)(dx+i dy) which is ∫_C(3x²–i y)dx+(y+i 3x²)dy. We write C as C₁+C₂ where C₁(t)=i t for t in the interval [–4Π,–2Π], and C₂(t)=0+2e^it for t in the interval [–Π/2,0]

• The C₁ integral
  Here x=0, dx=0, y=t, and dy=dt. Therefore
  ∫_C1(3x²–i y)dx+(y+i 3x²)dy=∫_–4Π^–2Πt dt=t²/2]_–4Π^–2Π=–6Π².
• The C₂ integral
  Here z=2e^it=2cos(t)+2i sin(t) for t in the interval [–Π/2,0], so that x=2cos(t), dx=–2sin(t) dt, y=2sin(t), and dy=2cos(t) dt. We translate:
  ∫_C2(3x²–i y)dx+(y+i 3x²)dy= ∫_–Π/2⁰I'm tired.

• Infinite series (especially power series)
• Line integrals

Infinite series
Everything is similar to what was done in calc 2. In the text this material begins on page 37. An infinite series is a sum of the form ∑_j=1^∞z_j. Associated to this "sum" is a sequence, the sequence of partial sums: s_n=∑_j=1ⁿz_j. If the sequence {s_n} converges and its limit is s, then we declare: the infinite series ∑_j=1^∞z_j converges and its sum is s. We're not computing infinitely many additions (!) -- we are trying to get the limit of a sequence.

Taking real and imaginary parts is linear, certainly for finite sums. This leads to a series version of the fact above.

Series Fact A complex series converges exactly when both of the real series gotten from the real and imaginary parts of the series terms converge. The sum of the complex series is the sum of the sum of the series obtained from the real parts plus i multiplied by the sum of the series obtained from the imaginary parts.

If r≠1, then a+ar+ar²+...+arⁿ=(a–arⁿ⁺¹)/(1–r). This is proved by multiplying things by 1–r and noticing the cancellations. Both a and r may be complex. Now if |r|<1, we know that lim_n→∞rⁿ=0. Therefore, since the finite sum is a partial sum of the whole geometric series, we have proved:

Convergent geometric series If a is a complex number and r is a complex number with modulus <1, then ∑_j=0^∞ar^j–1 converges, and its sum is a/(1–r).

Example
Every series in this course will be related to this result. Here is a silly but still amazing example. Suppose we want the sum of the series
    cos(Π/6)/3+cos(2Π/6)/3²+cos(3Π/6)/3³+cos(4Π/6)/3⁴+...,
that is, ∑_j=1^∞cos(jΠ/6)/3^j. I just asked Maple and got 7/146+(12/73)sqrt(3) as the answer. How did the darn program get this? Maybe I won't want to do the details myself, but I really would like to get an idea of the method. Please consider what follows.

Well, cos(jΠ/6)/3^j is the real part of e^i jΠ/6/3^j which is the same as (e^i Π/6/3)^j. This is a geometric series with first term e^i Π/6/3 and ratio e^i Π/6/3. The modulus of this is 1/3 and since this is less than 1, the series converges. Its sum is (e^i Π/6/3)/(1–e^i Π/6/3).

But e^i Π/6=sqrt(3)/2+i/2 (I know some friendly triangles!), so that (e^i Π/6/3)/(1–e^i Π/6/3) is (sqrt(3)+i)/2 on top and 3–(sqrt(3)+i)/2 on the bottom. If then one takes the real part and rearranges things, the result is, indeed, indeed, ... 7/146+(12/73)sqrt(3). Really. To me this is both impressive and absurd.

Absolute convergence
A series ∑_j=1^∞z_j converges absolutely if the series of real non-negative numbers ∑_j=1^∞|z_j| converges.

In calc 2, the following fact is used: if a real series converges absolutely, then it converges. Of course, almost immediately the instructor asserts that the converse is not correct, and the example exhibited is almost always the Alternating Harmonic Series, ∑_j=1^∞(–1)^j/j. This series converges by the Alternating Series Test (terms alternate in sign, decrease in value, and actually have limit 0). The "absolute" series corresponding to this is the Harmonic Series, ∑_j=1^∞1/j, which diverges (either use the Integral Test, or examine the series in bigger and bigger blocks of powers of 2, comparing it to a divergent series).

In fact, for complex numbers, absolute convergence also implies convergence. Why? Well, if ∑_j=1^∞|z_j| converges, then both ∑_j=1^∞|Re(z_j| and ∑_j=1^∞|Im(z_j| must converge (these are real series and I am using the real comparison test on them!). Since ∑_j=1^∞|Re(z_j| converges, then (using the calc 2 result!) ∑_j=1^∞Re(z_j) must converge. Similar logic for the imaginary part tells us that ∑_j=1^∞Im(z_j) must converge. But since the real and imaginary parts both converge, we then deduce that ∑_j=1^∞z_j converges. So:

Absolute convergence implies convergence for complex series:
If ∑_j=1^∞|z_j| converges, then ∑_j=1^∞z_j converges.

The Ratio and Root Tests are still valid for complex series. Their verification proceeds in much the same way as what we just did. So if we consider a series ∑_j=1^∞z_j of non-zero complex numbers (non-zero so I don't need to worry in what follows about division by 0) then:

• Root Test If lim_j→∞|c_j+1|/|c_j|=r exists, then when 0≤r<1, the original series converges absolutely and therefore must converge.
• Ratio Test If lim_j→∞|c_j|^1/j=r exists (the roots here are the usual non-negative roots of a non-negative real number), then when 0≤r<1, the original series converges absolutely and therefore must converge.

Here are some typical examples. They occur as solutions of certain differential equations but also in many other situations.

Classical example #1
∑_j=0^∞(–1)^jz^2j/[2^jj!]²
I am not making this up. This is J₀(z), the first Bessel function of order 0. It converges absolutely and therefore converges for all complex numbers, z. Such functions satisfy an ordinary differential equation: x²y´´+xy +(x²–ν²)y=0. Here ν=0.

The Ratio Test (which I always think of first when I see factorials) immediately (well, after some cancellations!) shows that this series converges for all complex numbers z. Here is a bit more information.

If c_j=(–1)^jz^2j/[2^jj!]² then |c_j|=|z|^2j/[2^jj!]² since the minus 1's drop out and the modulus filters through to the z's. Then:

|cj+1|     |z|2(j+1)/[2(j+1)(j+1)!]2     |z|2(j+1)[2jj!]2
------ = ------------------------- = ---------------------
 |cj|       |z|2j/[2jj!]2             [2(j+1)(j+1)!]2|z|2j

Classical example #2
The hypergeometric function ₂F₁(a,b;c;z) is a (not) well-known example of a collection of functions discovered(?) by Gauss and widely studied and generalized in the last hundred years. Indeed, even now these functions are objects of intense investigation. This is the traditional notation. Here it means the function which, near z=0, is defined by the following series:

    a·b       a(a+1)·b(b+1)        a(a+1)(a+2)·b(b+1)(b+2)
1+ ----- z + --------------- z2 + ------------------------- z3 + .... 
    1!c         2!c(c+1)                3!c(c+1)(c+2)

Let's look at an example, ₂F₁(1,2;3;z). What is this? So a=1, b=2, and c=3 and the series above becomes

    1·2       1(1+1)·2(2+1)        1(1+1)(1+2)·2(2+1)(2+2)
1+ ----- z + --------------- z2 + ------------------------- z3 + .... 
    1!3         2!3(3+1)                3!3(3+1)(3+2)

    1·2       1(2)·2(3)        1(2)(3)·2(3)(4)
1+ ----- z + ----------- z2 + ----------------- z3 + .... 
    1!3        2!3(4)             3!3(4)(5)

    2       2        2
1+ --- z + --- z2 + --- z3 + .... 
    3       4        5

In general, you'll see that a power series will have a radius of convergence and a disc of convergence with that radius. The series will converge (because it will converge absolutely) inside the disc. For z's not on the boundary but outside the disc, the series will diverge. For various reasons, we won't be too interested in what happens exactly on the boundary.

The book I mentioned
The "book" (reason for the quotes: there's not much of a plot, but lots of background characters!) I mentioned in class is Handbook of Mathematical Functions: with Formulas, Graphs, and Mathematical Tables edited by Milton Abramowitz and Irene A. Stegun. It is $26.37 from Amazon (Dover reprint), and has 1,046 pages. I bought it because I wanted to be respectable. I know less than .01% of the contents. You should read the reviews at Amazon before purchasing a copy. It has lots of "stuff" in it, almost surely more than you would want to know about a lot of things. It has many formulas. It is more fun than ... than ... well, maybe there's no good end to such a sentence. But once in a while reading it is both entertaining and informative.

Let's go on to discuss the other major tool in the course, line integrals.

Line integrals are integrals over curves in the plane. So I'll recite the standard curve vocabulary first.

Phrase	Proper definition	Possible picture
Curve	A curve is a function whose domain is an interval, [a,b], in R, and whose range is in C. If C(t) is the curve, we could write C(t)=x(t)+iy(t). In this course, these functions will generally be at least continuous.
Smooth (differentiable) curve	C(t)=x(t)+iy(t) should be differentiable, so that x(t) and y(t) are differentiable. Such a curve has a nice tangent vector at every point.
Piecewise smooth curve	The curve C(t) is certainly continuous. But additionally, we can divide the interval [a,b] into a finite number of pieces, and in each of these subintervals C(t) is smooth. This will allow us to take care of curves with finite numbers of corners, for example.
Simple curve	A simple curve is one which does not intersect itself (except maybe: see the next plus one definition). So if the domain of C(t) is [a,b], and if a≤s<t≤b, then C(s) is not equal to C(t).
Closed curve	"Head=Tail" -- that is, if the domain of C(t) is {a,b], then C(a)=C(b).
Simple closed curve	A closed curve which is also as simple as it can be(!). Here C(a)=C(b), and C(s)=C(t) with s<t implies either s=t or a=s and b=t.
I stopped here. I'll continue next time with the two definitions below and then do some examples and go on to line integrals.
Adding curves	Suppose C1(t) is a curve with domain [a,b] and C2(t) is a curve with domain [c,d]. If C1(b)=C2(c) (that is, C1's "Head"=C2's "Tail", perhaps) then the curve C(t) whose domain is [a,b+d–c] and defined by C(t)=C1(t) if t<b and C(t)=C2(t–b+c) (I hope!) is called the sum of C1 and C2. We will write C=C1+C2.
Reversing curves	If C(t) is a curve with domain [a,b], then the "reversal of C(t)", also called –C, will be the curve, let me temporarily call it D(t), whose domain will also be [a,b], defined by D(t)=C(b+a–t). –C is just C backwards. The picture is slightly weird (how should I have drawn it?) but I hope you get the idea.

I hope the darn formulas for C₁+C₂ and –C(t) are correct. I don't think I will ever be so formal again in this course. I will usually be rather casual. The curves in this course will all be piecewise differentiable, and, actually, I think every curve I'll consider will turn out to be sums of line segments and circular arcs. So bah to general theory!

Inequalities and complex numbers
So grading of your first formal homework assignment is ongoing. Even now, thousands of instructors are passing the papers from hand to hand, each evaluating every problem. Well, no: that's not quite correct. I've read solutions to problems A and B, and to the first textbook problem, and a grader will read the solutions to the other 5 textbook problems. It has been my experience with this course that the first real challenge to students is consideration of inequalities. Working with inequalities is the heart of analysis. The additional emphasis in this course is that we have complex algebra also on display. Let's review a few things about inequalities.

First, inequalities are really a shorthand way of dealing with positive quantities. People have recognized the following rules about positive numbers:

1. If w is a number, then w is positive, –w is positive, or w is 0. Exactly one of these alternatives is true. (We abbreviate the statement "w is positive" by the notation "w>0".)
2. The sum of positive numbers is positive.
3. The product of positive numbers is positive.

Now consider the complex numbers. The most interesting "new" number is i. If we have the rules above are correct, then i is 0 or i is positive or –i is positive. Certainly i is not 0 since its product with itself is –1, which isn't 0. If i were positive, then –1, the product of i with itself, would have to be positive. But we just saw that couldn't be true. If –i were positive, then the product of –i with itself would have to be positive, but that also is –1. All three alternatives seem to be impossible!

The correct conclusion is that there is no possible way to define a collection of positive complex numbers. There is no possible way to do inequalities with complex numbers themselves. Therefore, when I am grading homework problems or exam problems, if I see things like "5+6i> 2–i" or "w is a complex number and 5w+2>–4" then I can STOP READING. Any logical deductions which follow from such assumptions are dubious. They may or may not be true as logical statements but they certainly don't follow using any standard argument. You must convert complex numbers into real numbers in order to make valid statements using inequalities. The standard mechanisms use modulus or real part or imaginary part, but something has to be done. Bare statements with complex numbers and inequalities contain no information.

De Moivre The result (cos(θ)+i sin(θ))n=cos(nθ)+i sin(nθ) now becomes (eiθ)n=ei nθ which doesn't look so dramatic. I will usually write a complex number of modulus 1 in this course as ei(something real): this is easy and compact.

A^B
In this course, if A and B are complex numbers, we define A^B to be exp(A log(B)) which is usually written as e^A log(B). Note the small "l" in log. Since log can have infinitely many values (because arg can have infinitely many values), the symbol A^B can also may have infinitely many values.

A computation
For example, let's compute (1–i)³ⁱ. This should be e^{3i log(1–i)}. But log(1–i) is log(|1–i|)+i arg(1–i). Now log(|1–i|)=log(sqrt(2)), not so hard. But 1–i is in the fourth quadrant, and its Arg is –Π/4 so that arg(1–i) is –Π/4+2Πk where k is some integer (positive or negative or zero). Therefore (1–i)³ⁱ is e^{3i(log(sqrt(2))+i{–Π/4+2Πk})} which is e^{(3Π/4+6Πk)+3log(sqrt(2))i}.

This is e^(3Π/4+6Πk)cos(3log(sqrt(2)))+i e^{(–Π/4+6Πk)}sin(3log(sqrt(2))). When I ask Maple to compute (1–i)³ⁱ, the result is e^(3Π/4)cos(3log(sqrt(2)))+i e^(–Π/4)sin(3log(sqrt(2))) so it seems to use "our" Arg.

z²
Integer powers work out well with this definition of A^B. For example, z²=exp(2log(z))=exp(2{ln|z|+i(arg(z)+2Πk)})=exp(ln(|z|²)+i(2arg(z)+2Πk))=exp(ln(|z²|)+i arg(z²))=z² because exp is 2Πi periodic and modulus and argument work well with multiplication.

Back to abstraction ...
We need to discuss the foundational concepts of convergence and continuity. Convergence will be needed for both sequences and series. Continuity will be applied to functions. It is convenient to have this language. I will mostly be repeating things you can read about in section 1.4 of the text. Also, I'll just describe how to support complex computations with the corresponding real computations of calc 2.

If p is a complex number, then |Re(p)|≤|p| and |Im(p)|≤|p|.
This means that if the complex number p has small modulus, then both the real and imaginary parts of p are small. On the other hand, the Δ≤ (the Triangle Inequality!) tells me |p|≤|Re(p)|+|Im(p)|
so that if p has small real and imaginary parts, then the modulus of p is small.

Convergence of a sequence of complex numbers
This is on page 33. If {z_n} is a sequence of complex numbers, we will say that {z_n} converges to the complex number A if |z_n–A| gets small and stays small when n is sufficiently large.

If you wish, I could be more precise: given any positive number, ε, then there is a positive integer, N, so that if n≥N, then |z_n–A|<ε. The definition in this precision is quite famous, and took much effort by the 19^th century mathematical establishment to formulate. If you are interesting in investigating such statements, take 311, 411, 501, ... I will be content if you really believe "gets close and stays close".

Further, if {z_n} converges to A, then we also write lim_n→∞z_n=A or, sometimes, z_n→A.

You should notice that:

• If we consider the sequence {z_n} and we know that lim_n→∞z_n=A, then, since |Re(z_n)–Re(A)|≤|z_n–A| and |Im(z_n)–Im(A)|≤|z_n–A|, we know that the sequence of real numbers {Re(z_n)} converges with limit Re(A) and we know that the sequence of real numbers {Im(z_n)} converges with limit Im(A). So both of the real sequences converge, and the limits have the "correct" connection to the limit of the complex sequence.
• Again we start with a complex sequence {z_n}. From it we get two real sequences to consider: {Re(z_n)} and {Im(z_n)}. It may happen that both of these sequences converge. So suppose we know that lim_n→∞Re(z_n)=P and lim_n→∞Im(z_n)=Q for suitable real numbers P and Q. Then I claim that lim_n→∞z_n=P+i Q. Why is this true? Well, if |Re(z_n)–P| is small and |Im(z_n)–Q| is small then |z_n–(P+i Q)|≤|Re(z_n)–P|+|Im(z_n)–Q| so |z_n–(P+i Q)| is small, and lim_n→∞z_n=P+i Q.

Fact A complex sequence converges exactly when both of the real sequences gotten from the real and imaginary parts of the sequence elements converge. The limit of the complex sequence is the sum of the limit of the real parts plus i multiplied by the limit of the imaginary parts.

1. Suppose {z_n} converges. What can we say about {|z_n|} and {Arg(z_n)}? Well, |z_n|–|A|≤|z_n–A| and |A|–|z_n|≤|z_n–A| by our "reverse" triangle inequality. But this means ||z_n|–|A||≤|z_n–A| (no matter which one of |z_n| or |A| is larger!). But this means {|z_n|} converges and that its limit is |A|. So the convergence of one of the real sequences is just about clear. Must the sequence {Arg(z_n} also converge? Polar coordinates can sometimes be rather irritating, and this is an example. Remember that Arg of a complex number is the unique argument in the interval [–Π,Π). There must be trouble near the negative real axis. For example, consider the sequence given by z_n=–1+(–1)ⁿ(i/n). I know this sequence converges: its real parts are all –1; the imaginary part of the n^th term has absolute value 1/n, so those →0. The sequence converges and its limit is –1. The arguments certainly →1, because sqrt(1+{1/n²})→1. But what about Arg(z_n)? For n odd, (–1)ⁿ is –1, and z_n is in the lower halfplane, with Arg(z_n)→–Π. For n odd, the terms are in the upper halfplane, and since they approach –1 on the negative real axis, this subsequence of Arg's must →Π/2. This is annoying. In fact, if the limit of the complex sequence {z_n} is not on the negative real axis, then sequence of {Arg(z_n)} does "behave" and it has a limit which is Arg of the limit fo the complex sequence (this is seen using the calculus arctan function). I'm not interested in the details: I'll just try to avoid arguments with Args.
2. Suppose {z_n} is a sequence of complex numbers and we know both {|z_n|} and {Arg(z_n)} converge. Then since both sine and cosine are continuous, I know that |z_n|e^i Arg(z_n)=|z_n|cos(Arg(z_n))+i |z_n|sin(Arg(z_n)) converges. Wow! In fact, let me push this a bit more. Sometimes we can conclude that {z_n} converges only knowing about the convergence of one of the sequences {|z_n|} and {Arg(z_n)}. Can you see which one, and under what additional circumstance?
   I hope the assertion is the following: if we know that lim_n→∞|z_n|=0 then the sequence {z_n} must also converge and its limit will be 0. This is because the values of sine and cosine applied to various Args must be in [–1,1] so the product and sum will →0.

We'll have many sequences which converge to 0. Such sequences occur when we investigate infinite series. So if Q is a complex number with |Q|<1, the sequence {Qⁿ} converges to 0 since I know the real sequence {|Q|ⁿ} converges to 0. And, more weirdly, perhaps, would be defined by z_n=(3+4i)ⁿ/n!. This sequence converges to 0 because the real sequence {|z_n|} which is {5ⁿ/n!} has limit 0 (factorials grow much faster than any powers!).

Continuity
A continuous function transforms convergent sequences to convergent sequences. I mean: if {z_n} is any convergent sequence in the domain of the function f and if A, the limit of the sequence, is also in the domain of f, then the sequence {f(z_n)} must also converge, and its limit will be f(A). (There are other definitions of continuity, but, again, this won't turn out to be a major technicality in this course. [What will be, then?])

Almost every function in this course will be an example of a continuous function, so much so that it almost hurts in complex analysis to display functions which aren't continuous. I could "test" for continuous functions again by relying on real and imaginary parts. So z² is continuous because x²–y², the formula for Re(z²), and 2xy, the formula for Im(z²), are both continuous functions of x and y. And similarly e^z is continuous: consider e^xcos(y) and e^xsin(y). Rational functions of z are continuous in their domains, so 1/(z²–2) is continuous in all of C except ±sqrt(2). So that is not an example of failure of continuity.

A made-up terrible example
I could look at a piecewise-defined complex function. So we could have f(z) equal to z² for z with Re(z)≥0 (the closed right halfplane) and equal to z^–2 for z with Re(z)<0, the open left halfplane. This function is fine away from the imaginary axis. That is, if x=Re(z)≠0, I bet that f(z) is continuous. Why?

As before, if Re(z)>0, f(x+i y)=(x²–y²)+i(2xy) , and if Im(z)<0, we get (x²–y²)–i(2xy) multiplied by 1/(x²+y²)² because 1/z² is also (multiplying top and bottom by (z)²):
    (z)²/[(z²)(z)²].
The top is the conjugate of z² and the bottom is z²(z)² which is |z|⁴. Away from (0,0) this is fine.

What happens on the imaginary axis? The values are given by the z² formula, so we get (when x=0) –y². These values are also the limits "from the right". From the left, we'll get limits from the z^–2 formula if we stay away from z=0. If we set x=0 in that formula, the result is –y² multiplied by 1/y⁴ so the result is –y^–2. When is –y²=–y^–2? This equation asks for the limits from the left halfplane to be equal to the values and limits from the right halfplane. This happens when y=±1. So there is continuity at ±i. By the way, f(z) is also not continuous at z=0, since z²→0 but z^–2 does not →0 as z→0.

To summarize for this f(z): continuous if Re(z)≠0, and also at ±i. f(z) is not continuous at any other points.

I don't believe we will need to discuss such an example again in this course.

There will be two examples: the most important function in the world (the exponential function) and its close relatives (this will have some surprises), and the much more ordinary linear function (but some novelty will even occur there). I am interchanging the order of two sections of the book (1.4 and 1.5) with this presentation.

There are many ways to introduce the exponential function. Here I'm not going to ignore the considerable background you all have, so what follows is not mathematically "pure", which a number of people prefer. Purer approaches are found in many books.

The single most important function
What should the exponential function be?

1. Certainly exp(0)=1.
2. I think exp(z₁+z₂)=exp(z₁)exp(z₂) for all possible complex z₁ and z₂.
3. When we can differentiate officially, certainly exp should be differentiable, and exp´(z) should be exp(z).

The textbook asserts that

What is and what should be the exponential function
Here is one heuristic way of seeing why the definition is "correct". Here "heuristic" means "allowing or assisting to discover." It turns out (you will see) that everything I'm doing can be made (terrifically!) rigorous. So this is a way of seeing why people use this strange definition.

If exp(z)=rxp(x+i y)=exp(x)exp(i y), then we just need to understand exp(i y). If f(y)=exp(i y), then f(0)=exp(0)=1. Also, if we believe in the chain rule, f´(y)=exp´(i y)i, and then f´´(y)=exp´(i y)i ²=–exp´(i y). So f(y) is a function which is minus its own second derivative, and we have the initial conditions f(0)=1 and f´(0)=i . Well, we know from our differential equations course that any function which is minus its second derivative (the Vibrating Spring -- one of the first o.d.e.'s studied) is a linear combination of sine and cosine, so exp(i y) must be Acos(y)+Bsin(y) for some A and B. The initial conditions allow us to conclude that A=1 and B=i.

How to compute it
This description is made of sums and products of well-known functions which are certainly continuous, and therefore it is nice and continuous, also. I can compute e^1.2+.7i because I can compute e^1.2 and cos(.7) and sin(.7) and then combine them appropriately. So e^1.2+.7i=e^1.2cos(.7)+i e^1.2sin(.7)≈2.539365489+2.1388780451, according to Maple. And, by the way, when I request the argument of this complex number from Maple, the result is ... 0.7000000000. Should this be expected?

Periodicity
The most shocking thing about the exponential function is that it is periodic. Since e^iy is cos(y)+i sin(y), changing y by an integer multiple of 2Π won't change e^iy. Therefore, for all z, e^z+(2Π)i=e^z. (!)

Solving e^z=w
If e^z=w, then e^x+i y=s+i t, so
    e^xcos(y)=s and e^xsin(y)=t.
Therefore we deduced that w cannot be 0 (or else either e^x=0 [impossible!] or both sin(y) and cos(y) would be 0 [impossible, since sin²+cos²=1]). If w is not 0, then we got solutions of e^z=w: x=ln(|w|) and y=arg(w). For non-zero w, (not surprisingly) a solution of e^z=w is called log(w). Frequently people want a specific solution and for that we need to get some notation and designate a specific value of arg.

From arg to Arg
If z is a non-zero complex number, then arg(z) is any real number which makes the equation z=|z|(cos(arg(z))+isin(arg(z))) correct. A simple picture should convince you that there are such numbers. In fact, there are infinitely many numbers which can serve as arg(z) since sine and cosine are 2Π-periodic. arg(z) therefore specifies an infinite collection of numbers for each non-zero z. There is a unique number which can serve as arg(z) inside any interval of length 2Π. That is, if you specify A, and then we search the interval [A,A+2Π), we will find exactly one number which will be a valid arg(z). Most people would like to know a specific value of arg in order to compute efficiently. But, hey, we could specify A=10¹⁰ or A=–117. Neither of these are used. The most commonly used values of A are 0 and  Π. Our text uses A=–Π, and calls the unique value of arg(z) which is in the interval [–Π,Π), Arg(z), with a capital A.

So while arg(1+i) is the collection of numbers {Π/4, 9Π/4, –7Π/4, 17Π/4, –15Π/4, ...} the number Arg(1+i) uniquely specifies Π/4. And Arg(-1-sqrt(3)i) is uniquely -2Π/3 (if I am imagining the triangle correctly!). The only numbers which have Arg(z)=0 are the positive reals.

Level curves of Arg
The picture below shows some level curves of Arg. Notice that the origin has no value for Arg. Each of the level "curves" is a ray, a half line, starting from the origin (but not including it!). So Arg is a peculiar function. The last level set is actually empty!

The specific A chosen in the text (so [-Π,Π) contains all of Arg's values) is so that Arg will be continuous near points on the positive real axis.

Suppose A is a point on the positive real axis. Then near A, Arg will vary continuously. Points near A in the upper halfplane will have Arg values near 0 and positive, and those near A in the lower halfplane will have Arg values near 0 and negative. Suppose, though, we try to move farther away. Look at the two parts to the right. On the Upper curve, the values of Arg will increase (except around the loop!) and especially as the Upper curve points near B (on the negative real axis), they will increase to Π. What happens on the lower loop? There the values of Arg will descrease, and as the points move (?) towards B, the values of Arg will tend to -Π. So Arg is actually not continuous at B. Depending on the direction of approach, there may be a jump of + or –2Π in Arg's value. So:
Arg is not continuous at any negative real number, and at such a number, Arg has a jump discontinuity of magnitude 2Π.

From log to Log
Again, for non-zero w, a solution of e^z=w is called log(w). If y=Arg(w), then the solution is called Log(w) (note the capital letter!). The different formulas for log give what are sometimes called branches of log. In this jargon, Log would then be called the principal branch becuase it has the standard real information, the most familiar, included in it.

The complex exponential function has many pleasant and some irritating aspects. It is a bit difficult to draw a true graph of this function. The domain points are in C which is R², as are the range points. Therefore the graph, the pairs of points (z,e^z) would be elements of R⁴, and I don't know any neat ways of showing four-dimensional graphs. (Many attempts of been made. For example, we could sketch (x,y,u(x,y)) and change this standard graph of a surface in R³ to the graph (x,y,v(x,y)) by some geometric process. Or one could "draw" the graph of u(x,y) and put colors on the graph according to the values of v(x,y). You can easily find these methods and others on the web. I'll stick to more ordinary methods here.

The most interesting aspect of the function which will be shown in its mapping properties is that exp is 2Πi periodic. This means that exp is infinite-to-1. Every non-aero complex number w has infinitely many z's with exp(z0-w, and these are all separated by 2Πi.

I'll have a complex plane representing the domain to the left, and then a range complex plane to the right. I'll try to sketch objects on the left in the domain and then sketch the corresponding objects on the right, objects which "correspond" under the exponential mapping.

One horizontal line Consider the horizontal axis. We could image a bug moving on the axis in the domain at uniform velocity, from left to right. Here y=0, and x goes from –∞ to +∞. The image in the w-plane is excos(0)+i exsin(0) which is ex. Notice that this is a ray, a halfline, whose image covers the positive x-axis. There is a further difference: kinetic. The image of uniform motion is no longer uniform motion. Near "–∞" the image bug moves very slowly, and as the image goes from left to right along the positive x-axis, the image bug moves ever faster, exponentially faster.

One vertical line Now consider the vertical imaginary axis, where x=0, and a bug crawling at uniform velocity from –∞ (in y) to +∞ (in y). Now with x=0, the image is e0cos(y)+i e0sin(y). So the image is the unit circle, described in the positive (counterclockwise) fashion, with the circle drawn at uniform velocity (1 radian per timestep).

Other horizontal lines If we look for the image of a horizontal line which is just slightly above the x-axis, what will we get? Here we have excos(y)+i exsin(y) with y a small positive number. In fact, in terms of vectors, this is all positive (ex) multiples of the unit vector [cos(y),sin(y)]. So the image is again a halfline, a ray, emanating (originating?) from 0 at angle y to the positive x-axis. You could imagine as the horizontal lines go up or go down, the images will rotate around the origin.

Other vertical lines If we look for the image of a vertical line which is to the right of the y-axis, what will we get? Here we have excos(y)+i exsin(y) with x a positive number. In fact, this describes a circle whose center is the origin, and whose radius is ex, always a positive real number. You could imagine as the vertical lines go left or right, the images will all be circles centered at the origin of varying radii.

An oblique line
We even briefly considered an oblique line, and decided that its image was a spiral. So if y=mx, say, with m>0, then the image under exp of this is exp(x+i my)=e^xcos(my)+i e^xsin(my). This is a spiral. somehow "interpolating" between a circle (the image of a vertical line) and a ray from the origin (the image of a horizontal line). To the right is a Maple of part of one of those spirals with m=8.

Although Maple has a complexplot command (part of the plots package) this is what I did to create the picture shown. First, from y=8x I decided that I needed to understand exp(x+8x i). So I computed:
     exp(x+8x i)=exp(x)exp(8x i)=e^x(cos(8x)+i sin(8x))= e^xcos(8x)+i e^xsin(8x))
and from this used the Maple command
plot([exp(t)*cos(8*t),exp(t)*sin(8*t),t=-0.01..2],numpoints=100,thickness=2,color=black);
(I increased the default of numpoints from 50 because I wanted a smoother curve.

Strips mapped by exp
We also saw how strips were mapped by exp. A horizontal strip which is less than 2Π high will be mapped into a wedge of the plane, with the corner of the wedge at 0. exp exactly matches the points in the horizontal strip 1-to-1 with points in the wedge.

A vertical strip will always be mapped onto an annular region, but this mapping is infinite-to-1.

Finally, one last picture which attempts (poetically? qualitatively?) to show you what exp does to the plane. We start on the left with an R². Then we pick it up and a hand, infinitely far to the left (?), "squeezes" that side to a point. Then we expand this fan over and over itself (?) infinitely many times and finally we project the resulting bunch of knitted-together (?) discs onto another copy of R². This is the geometry of the exponential mapping on a big-scale level.

Some finer features which turn out to be very important in applications are not at all clear in this presentation. For example, curves in the left-hand, domain, plane are mapped to curves in the right-hand, range, plane so that the angles of intersection of the curves do not change! This is not obvious, and we will study these angles in detail later.

Here, in italics, are notes from a previous instantiation of this course describing the experiment with logs that I did, this time, without student participants.

Three of the wonderful students in the class volunteered (always better than the truth, which is closer to were dragooned) to draw some pictures with me. I only show the results of three of the students. I set up the exponential function mapping from a domain copy of C to a range copy of C. Then I stood at the range copy, with my chalk at w=1, which I also called HOME. I asked each of the students to choose a value of log(1) different from 0. Since this is a complex analysis course, there were certainly enough distinct values to choose from. Each student could have their own value of log(1). In my head I saw a sort of mechanical linkage reaching backwards from my 1 to their choices of log(1). The linkage would "automatically" respond to any changes of position in my location by moving to nearby places whose value of exp would be where my chalk moved to. Sigh. There's no pathology in this course: things will be continuous, things will move nicely, etc. Much more subtle problems will be met.

The first trip Then I carefully and slowly (well, almost) moved my chalk from HOME in a simple closed curve. The curve was made up of circular arcs centered at the origin and pieces of rays which started at the origin. I used these curves because last time we had previously computed how vertical and horizontal lines in the domain were mapped. The inverse images of my "trip" from HOME to HOME were carefully drawn. They were congruent, because exp is 2Pii periodic. Each started from each student's choice of log(1) and returned to that choice.

The second trip Well, now I left HOME and traveled in a circle around 0, and returned to HOME. The students attempted "follow" me in the domain of exp, and there was some controversy. If we tried to make everything continuous, then, even though my path in the range was a simple closed curve, each student draw a path going from one choice of log(1) to another choice of log(1). This is very, very mysterious. I drew a few more paths in the range and mentioned what would happen in the domain, and said we would need to explain all this is great detail later. This is one of my goals in the course.

Linear mappings
We know (we have linear algebra knowledge) that a random linear mapping from R² to R² is represented by a matrix,

( a  b )
( c  d )

The surprise, which will turn out to be profound, is that there is a serious restriction. We want to study in this course only linear mappings which occur as a result of multiplication by a complex number. So if z=x+i y, and Q=α+i β, then we would like to look at only mappings coming from z→Qz (complex multiplication!). Now what happens algebraically? Look:
    (x,y) corresponds to z=x+i y → Qz=(αx–βy)+i (αy+βx) which corresponds to (αx–βy,αy+βx)
This last vector is the result of using the 2-by-2 matrix

( α  –β )
( β   α )

Nouns?
By "nouns" I mean here some of the standard terminology which is used in "analysis" -- the grownup word for calculus and all of its extensions. There will be a string of definitions, maybe punctuated by some examples and some questions. The definitions are just to give us enough common words so we can discuss usefully the phenomena (!) of the course. I will not concentrate on the logical intricacy of the definitions (we have several other courses doing that).

Dr(p) The open disc of radius r whose center is the complex number p.	Here p is a complex number and r is a positive real number. Then Dr(p) is the collection of complex numbers z with |z-p|<r
An open set of complex numbers	Suppose U is a collection of complex numbers. Then U is open if for every number p in U, there is some positive real number r (which may depend on p!) so that Dr(p) is entirely contained in U. You've got to be able to put a disc inside U for every p in U.
Boundary point	The complex number q is a boundary point of a set W of complex numbers if every disc of positive radius r centered at q, Dr(q), contains at least one point in W and at least one point not in W. For the sets shown to the right, t is in U and is not a boundary point of U, because the disc shown is entirely inside U and has no points of V. The point s is a boundary point of both V and U since every disc of positive radius has points in V and points in U.
Polygonal curve	A polygonal curve is a finite number of line segments L0, L1, L2, ..., Ln, so that the end of Lj is equal to the beginning of Lj+1 for 0<j<n.
Connected open set	An open set is connected if every pair of points in the set can be joined by a polygonal line which is entirely inside the set. Again, every pair of points must be joined by some polygonal line totally inside U.

Now let us consider some examples.

The set & its owners	Is it open?	If it is open, is it connected?	Describe all the boundary points of the set	Picture of the set
D1(0) This is the open unit disc: those z's with |z|<1. Owned by: Mr. Glass & Mr. Breuer.	The set is open (needs Δ≤ to check this: for a disc centered at p use as radius 1–|p|, or look at the picture).	Yes, connected. Any two points in the disc have a line segment connecting them which lies entirely inside the disc.	The boundary points satisfy |z|=1. Any disc of positive radius centered at such a point has points both in the set and not in the set.
A set is convex if, whenever two points are in the set, the line segment containing the points is entirely within the set. So a circle (the boundary of a disc!) is not convex, but a line segment itself is convex. The unit disc is an example of a convex open set, and convex open sets are always connected (the polygonal curve is just one line segment!)
z's with |z|≤1 Sometimes called the closed unit disc. Owned by: Ms. Janiszewski & Mr. Sirisky.	This is not open. Consider the point p=1, in this set. Every disc of positive radius centered at 1 has points on the real axis to the right of 1, and therefore such a disc is not contained in the set.	Does not apply.	Again, the boundary is |z|=1. The verification is the same as the previous example.
z's with 1<|z|<2 This is called an annulus (the region between two concentric circles). Owned by: Mr. Russo & Mr. Zhang.	This is an open set.	Yes, this open set is connected. However, the connecting polygonal line may need more than 1 segment to go from one point to another. (I think it may need as many as 3 segments -- can you think of a connected open set which might need more?)	The boundary has two "pieces": one of all the z's satisfying |z|=1 and the other of z's satisfying |z|=2.
z's with 1<|Re(z)|<2 Owned by: Ms. Addabbo & Mr. Laflotte.	This is an open set. If p is in the right-hand strip, say, then 1<Re(p)<2, then a disc of radius min(2–Re(p),Re(p)-1) will be inside the strip. Dics can be created for points in the left-hand strip similarly. strip is	No. This set is not connected. For if it were, then the function Re(z) would have to be continuous when restricted to the polygonal line. If we try to "connect" –1.5 and 1.5, the function (by the one-variable Intermediate Value Theorem) would be 0 somewhere on the polygonal line, and no point in the set can have Re(z)=0. The polygonal line cannot be contained in the set.	The boundary is 4 straight lines :Re(z)=+/–1 and Re(z)=+/–2.
z's where z3/(z2+3z+2) is defined Owned by: Mr. Dyson & Mr. Yeager.	We'd better throw away z's where z2+3z+2=0. We can factor polynomials in a field to get roots, so z2+3z+2=(z+2)(z+1) indicates that –1 and –2 should be thrown out. The resulting set is open (if p is in it, take a radius of min(|p–(–1)|,|p–(–2)|) to get a disc entirely in the set).	This set is connected. Draw some vertical and horizontal lines through any pair of points in the set, and once they are away from the real axis, they can be "connected".	The boundary consists exactly of the two points –1 and –2, because a disc around one of those points will aways contain points in the set and also will contain the center of the set, which is not in the set.
Associated question Is C "minus" a finite number of points open and connected? (That is, everything except those finitely many complex numbers.) What are the boundary points of this set? Answer Yes, such a set is open and connected, and the boundary is just the finite number of points which were thrown out. Therefore, for example, the domain of a rational function (quotient of polynomials) will be a connected open set.
All z's with 1≤Re(z)<2 and 1<Im(z)<2 Owned by: Mr. Leung & Mr. Mohamed.	This is not as open set. Just look at the point z=1+1.5i. No disc of positive radius centered at 1+1.5i is entirely in the set.	Does not apply.	The boundary is 4 line segments: Re(z)=1 and 1≤Im(z)≤2, Re(z)=2 and 1≤Im(z)≤2, 1≤Re(z)≤2 and Im(z)=1, and 1≤Re(z)≤2 and Im(z)=2.

Comments on strategies
To verify in detail that a set is open, you must show that if z is any point in the set, then there is at least one positive radius r so that D_r(z) is in the set.

To verify in detail that a set is not open, you must shown that there is at least one z in the set so that if r is any positive radius, then D_r(z) always has points not in the set (it is not contained totally in the set).

So the logic is a little twisty: the quantifying phrases "all" and "there is at least one" play a part in both verifications. Such logical intricacies will not be a large part of this course -- we have other sorts of intricacies!

Question α for Mr. Marcus & Mr. Ronan:

1. Suppose an open set U has the property that for all p in U, D₅(p) is contained in U. What is U?
   Answer First a little bit of logic: the empty set (the set with no complex numbers in it at all!) does satisfy this specification, because of the vacuous nature (no requirement!) of the hypothesis. The usual symbol for this set is ∅. But what if U is not empty? Suppose p is in the set and q is some other complex number. Consider the line segment joining p and q. "Walk" along that line segment with steps of length 4, say (the key is that 4<5 here!). All of the points in each segment, starting at p, must be in U because the discs successively are all in U. So eventually we see that q is in set also. So actually, U must be all of C!
2. Suppose an open set U has the property that for all p in U, D_1/17(p) is contained in U. What is U?
   Answer U will either be the empty set or all of C. The reasoning is sort of the same, using steps of length less than 1/17.
3. Suppose an open set U and there is a positive real number r so that for all p in U, D_r(p) is contained in U. What is U?
   Answer Again, U is either empty or all of C.

Question β for Mr. Cantave & Mr. Khor:
Must the intersection of two connected open sets be connected? If the answer is yes, briefly explain why. If the answer is no, give some useful examples showing what can happen.
Answer The intersection of two open connected sets may be connected. For example, the intersection of two open discs is an open connected set. But the intersection may not be connected, for example, consider the picture to right. The two "blobs" pictured are open, and their intersection has two pieces. This somewhat simple picture will turn out to have important computational consequences.

Question γ for Mr. Sherman & Ms. Koshibe:

(Many 's not shown!)

1. Suppose S is the sequence of numbers 1, 1/2, 1/3, 1/4, 1/5, .... (this is an infinite collection of numbers). Suppose W is all complex numbers except those in S (the complement of S). Is W open? What are the boundary points of W?
   Answer Well, the problem is at 0. 0 is in the set W, but every disc of positive radius centered at 0 will have some of the 1/n's in it. So W is not open. The boundary points of W are all the points of S (just as in the simpler example with two points removed in the table able) and also the point 0, since there are also points of W in any disc of positive radius.

   (Many 's not shown!)

2. Suppose T is the following set: the sequence of numbers 1, 1/2, 1/3, 1/4, 1/5, .... (this is an infinite collection of numbers) and also the number 0. Suppose Q is all complex numbers except those in T (the complement of T). Is Q open? What are the boundary points of Q?
   Answer We seem to have removed the problem! If a point is far away from the sequence T and is not 0, then there is certainly a disc entirely contained in Q with center at that point. So the set Q is open. Its collection of boundary points remains the same as in the previous example: all the points of S and also the point 0.

Notice the examples just presented each involve taking an infinite number of points out of C and then inspecting the result. If we take only finitely many points away, then (see the example above analyzed by Mr. Dyson & Mr. Yeager) the result is always open and connected. Here we see that the "geometry" of the infinite number of points allows a more varied result. The set might or might not be open. I can tell you (this is not obvious!) that if the set is open, then it must be a connected open set.

A closed set is one which contains its boundary.

Here is a theorem on page 25 of the text: Theorem A set U is open if and only if if contains no point of its boundary. A set K is closed if and only if its complement (those complex numbers which are not in K!) is open. The proof is on page 25, and I will sometimes use this result but I don't want to go through the proof which is logical tail-chasing. (I am not declaring that the proof is silly, just that it is not in the mainstream of the concerns of this course!) If you like this sort of reasoning, you can take Math 311 or Math 441.

Which of the sets we have investigated are closed? Which are neither open nor closed?

Answers

• The open unit disc, the set of Mr. Glass & Mr. Breuer, is open and is not closed.
• The closed unit disc, the set of Ms. Janiszewski & Mr. Sirisky, is closed and is not open.
• The annulus, the set of Mr. Russo & Mr. Zhang, is open and is not closed.
• The double infinite strips, the set of Ms. Addabbo & Mr. Laflotte, is open and is not closed.
• The "doubly punctured plane", the set of Mr. Dyson & Mr. Yeager, is open and is not closed.
• The weird square, the set of Mr. Leung & Mr. Mohamed, is not open and is not closed. (Nomination for the worst math terminology in English and in the other languages I know: "open" and "closed": I know sets which are both [below] and I know sets which are neither!)
• In question α, analyzed by Mr. Marcus & Mr. Ronan, the empty set and the whole complex plane are both open and closed.
• In question β, analyzed by Mr. Cantave & Mr. Khor, I can tell you that the intersection of two open sets will always be open (but the intersection of infinitely many may not be!). The intersection of two closed sets is always closed, and, in fact, the intersection of any number of closed sets is closed. If you know de Morgan's laws (how to manipulate sets) then you can deduce what happens to unions of these sets. We will not need such results here.
• In question γ, analyzed by Mr. Sherman & Ms. Koshibe, the set W is not open and it is not closed. The set Q is open and is not closed.

Simple ? stuff
I'll repeat a few geometric definitions, stated in "complex language". If z=a+bi, then a=Re(z), the real part of z, and b=Im(z), the imaginary part of z. Also, we know some polar language: z=r(cos(θ)+i sin(θ)) where we really know θ only up to a multiple of 2Π and |z|=sqrt(a²+b²). (There is a unique value of arg(z) in any half-open interval of length 2Π such as [0,2Π) or (-Π,Π] but different situations lead to different choices!) If z=a–bi, the complex conjugate of z, then zz=a²+b² so that |z|²=zz. Please also note that if z=r(cos(θ)+i sin(θ)) then z=r(cos(θ)–i sin(θ))=r(cos(–θ)+i sin(–θ)), and therefore the modulus of z, |z|, is the same as |z| and the argument of (or, one argument of!) z is –arg(z). We showed last time that, with multiplication, the modulus multiplies and the arguments add. Therefore |z||z|=|z| |z|=|z|² and arg(|z|²) is arg(z)+arg(z)=arg(z)–arg(z)=0, so that |z|² is real and non-negative (this is reassuring!). Therefore |z|²=z z. And notice that |zw|², which must be (zw)zw, is also equal to |z|² |w|²=zzww. Therefore zw=z·w.

The triangle inequality
Much of this course will be devoted to getting precise and elaborate formulas for complicated things. The most amusing aspect of this endeavor is that it will depend on getting very sloppy estimates of fairly simple things. The basic sloppy estimate for most of analysis is the triangle inequality:

It is almost always useful to understand things in more than one way. Let's try to verify the triangle inequality algebraically. |z+w| has a square root which I will try to avoid, so I'll begin with its square: |z+w|²=(z+w)(z+w). But z+w is the same as z+w (check this either geometrically by flipping z+w or algebraically by changing the sign of the second coordinate in the two dimensional vectors). Let's try again:
   |z+w|²=(z+w)(z+w)=(z+w)(z+w)=zz+zw+wz+zw=|z|²+|w|²+wz+zw.
I will admit that one of the reasons for going through this verification is so that you see some of the algebraic tricks we will be using during most of the course. I need to analyze zw+wz. Well, we do know already that Q·P=Q·P. Another useful fact is conjugating a complex number twice gets back the original number (bar-bar is the identity). You can see this geometrically by flipping twice across the x-axis, or algebraically by – · –=+.

The "cross-term" is the sum of zw and its complex conjugate. But:
   Re(P)={1/2}(P+P} and Im(P)={1/2i}(P–P).
Therefore zw+wz is actually 2Re(zw). But:
   Re(P)≤|Re(P)|≤|P| and Im(P)≤|Im(P)|≤|P|

So:
|z+w|²=|z|²+|w|²+wz+zw≤ |z|²+|w|²+2Re(zw)≤ |z|²+|w|²+2|z|·|w|=(|z|+|w|)².
Taking square root (just square root of non-negative reals!) gets us the triangle inequality.

Some numerical estimates
Suppose we have P(z)=z⁵–(2+i)z³+9. How big can |P(z)| be on the circle of radius 2 centered at 0 (that is, where |z|=2)? Here we go:

|P(z)|=|z⁵–(2+i)z³+9|≤ |z⁵|+|–(2+i)z³|+|9|=|z|⁵+|2+i|·|z|³|+9=2⁵+sqrt(5)2³+9=41+8sqrt(5).

Can we get an underestimate? Here is an important variant of the triangle inequality:

We can now conclude that the values of P(z) for |z|=2 must lie inside an annulus centered at 0. That is, if |z|=2 and w=P(z), then r≤|P(z)|≤R where R=41+8sqrt(5)≈58.9 and r=23–8sqrt(5)≈5.1. The picture shown is not true to scale. The r and R and 2 are not in correct proportion. Notice I am not asserting that every w inside that annulus is a value of P(z) for |z|=2. That is false. In fact, the image of the circle is a curve inside the annulus. We'll see that the curve in fact twists 5 times around the center of the annulus.

I'll use this background color for material I didn't do in class because it wasn't convenient (some computations or pictures, for example!) or because I didn't have time, but which you might find interesting. What does it really look like? Here is a Maple picture of the image of |z|=2 under the mapping z→P(z): the black curve. The red circles are circles of radius 5.1 and 58.9 centered at the origin. The curve is inside the predicted annular region. It is certainly not always true that the methods outlined always will give such a tight fit. I think we are just lucky here because P(z) is fairly simple!

Quick, let's try again: consider Q(z)=z²–7z. Then |Q(z)|=|z²+7z|≤|z²|+|7z|=|z|²+7|z| and this is at most 18 if |z|=2. What about an underestimate? Some thought will show that we get information in this case if BIG is 7z (which has modulus 14) and LITTLE is z² (modulus 4). So 14–4=10 is an underestimate. And therefore 10≤|Q(z)|≤18 if |z|=2.

We can compound the achievement (if we call it that!). We could consider the rational function R(z)=P(z)/Q(z)=

z5–(2+i)z3+9
------------
   z2–7z

• To get an underestimate of a quotient, take the min of the top and the max of the bottom.
• To get an overestimate of a quotient, take the max of the top and the min of the bottom.

Some algebra
We know that if z=r(cos(θ)+i sin(θ)) and z=s(cos(λ)+i sin(λ)), then zw=rs(cos(θ+λ)+i sin(θ+λ)). Careful use of this formula will allow use to "explicit" solutions to certain rather simple polynomial equations. Later in this course, we'll see that all polynomial equations have solutions, although there won't be nice formulas for them.

Let's try to solve z³=5i. The modulus of 5i is 5 and its argument (or, better, an argument) is Π/2. If z=r(cos(θ)+i sin(θ)) then z³=r³(cos(3θ)+i sin(3θ)). When will this be 5i? Certainly the modulus information is carried by r³, so that if z is a cube root of 5i, then r³=5 and r must be 5^1/3. By 5^1/3 I mean precisely the unique positive real number whose cube is 5. (That's about 1.71.) What about the argument? I know that cos(3θ)+i sin(3θ))=0+1i, which means that θ is some number with cos(3θ)=0 and sin(3θ)=1. Let's think: 3θ should be Π/2 (plus possibly some multiple of 2Π) or 3Π/2 (plus possibly some multiple of 2Π). If 3θ is the first alternative, then θ must be
   Π/6 or
   Π/6 + 2Π/3 or
   Π/6 + 4Π/3.

If 3θ is the second alternative, then
   Π/2 or
   Π/2 + 2Π/3 or
   Π/2 + 4Π/3.

But we also need sin(3θ)=1. Since sin(3{Π6})=1 and sin(3{Π2})=–1, the solutions are only those resulting from the first alternative. We have deduced that solutions of z³=5i include these numbers:
   5^1/3(cos(Π/6)+i sin(Π/6))    5^1/3(cos(Π/6+2Π/3)+i sin(Π/6+2Π/3))    5^1/3(cos(Π/6+4Π/3)+i sin(Π/6+4Π/3))
We can even write these in rectangular form since I happen to know that sin(Π/6)=1/2 and cos(Π/6)=sqrt(3)/2. The solutions we have found can also be written as:
   5^1/3(sqrt(3)/2)+5^1/3(1/2)i; –5^1/3(1/2)+5^1/3(sqrt(3)/2)i; –5^1/3(sqrt(3)/2)–5^1/3(1/2)i

Some questions: are there any other solutions to the equation z³=5i? What is the significance (if any!) of these solutions?

Let me answer the first question. If we are working in a field, then most of high school algebra carries over (anything relying on addition, multiplication, subtraction, division). In high school we saw that if P(UNKNOWN) is a polynomianl, and if α is a root (so P(α)=0) then P(UNKNOWN)=(UNKNOWN–α)Q(UNKNOWN) where Q(UNKNOWN) is a polynomial of degree one less than the degree of P(UNKNOWN). This leads to the realization that a polynomial can have only as many distinct roots as its degree. Since z³–5i has degree 3, it can have only 3 distinct roots, and we've found 3 distinct roots. So there aren't any more.

The picture is incredibly full of information. It shows 5i, and then the 3 cube roots are displayed. Notice please that they form the vertices of an equilateral triangle whose center is at the origin, and whose radius is 5^1/3. The angular "offset" from the positive x-axis of a vertex of this equilateral triangle is one-third of the argument of 6i.

I hope that this discussion will persuade you to accept my declaration (!) of all solutions of the equation zⁿ=A where A is any complex number (well, if A=0, then the solution(s) are all 0, so maybe I'd better write "A is any non-zero complex number") and n is a positive integer.

The geometry of z when zⁿ=A if A≠0
Here is how to display the n solutions of the equation zⁿ=A for a non-zero A. First, draw a circle of radius |A|^1/n (a unique positive real number). Now draw an equilateral polygon with n sides inscribed in this circle. Here inscribed means that the corners or vertices of the polygon are on the circle and the sides are all inside the circle. For precision, put one of the vertices on the positive real axies. Then the polygon should be offset or rotated in the positive, counterclockwise direction by the argument of A divided by n. The vertices of the resulting object are the distinct n^th complex roots of A.

Examples
Probably the most important collection of examples occur when A is just 1. The roots are called the n^th roots of unity. Below is a picture of these roots for n running from 2 to 6. Since arg(1)=0, the rotational offset is 0, and one vertex of the regular polygon determined by the roots will always be 1.

O.k.: let me try to do more specific computation in one last example: z⁴=–1+i. Here 1+i has modulus sqrt(2) and (one value of) argument 3&Pi/4. So what are the fourth roots of 1+i? Since |1+i|=sqrt(2), I know that the fourth roots all must like on the circle of radius 2^1/8 centered at the origin. What about their arguments? Well, they will be "separated" by 2Π/4 and will "start" at (3Π/4)/4 which is 3Π/16. So here are the 4^th roots of 1+i:
A=2^1/8(cos(3Π/16)+i sin(3Π/16)); B=2^1/8(cos(11Π/16)+i sin(11Π/16)); C=2^1/8(cos(19Π/16)+i sin(19Π/16)); D=2^1/8(cos(27Π/16)+i sin(27Π/16)).
In the picture to the right, the circle is supposed to have radius 2^1/8.

Comments and objections
I didn't really compute them, did I? I think I did, but what if you wanted a rectangular description of them? Let's try the first one and try to get some numbers for cos(3Π/16)+i sin(3Π/16). I know cos(3Π/4): that's –1/sqrt(2). But I also know that cos(2w)=2(cos(w))²–1. So (taking w=3Π/8) –1/sqrt(2)=2(cos(3Π/8))²–1. This has roots (1/2)sqrt(2–sqrt(2)) and –(1/2)sqrt(2–sqrt(2)). I'll take the positive root since cos(3Π/8) is positive (0<3Π/8<&Pi/2). Then another half angle computation leaves me the choice between (1/2)sqrt(2+sqrt(2–sqrt(2))) and –(1/2)sqrt(2+sqrt(2–sqrt(2))) so I choose (1/2)sqrt(2+sqrt(2–sqrt(2))), and this is cos(3Π/16). Then sin(3Π/16) turns out to be (1/2)sqrt(2–sqrt(2–sqrt(2))). The first (?) root in the list above is therefore 2^–7/8sqrt(2+sqrt(2–sqrt(2)))+ 2^–7/8sqrt(2–sqrt(2–sqrt(2)))i. Maple tells me that the fourth power of this is indeed –1+i. (Wow! Of course this is rather special reasoning, using crazy identities which in fact are best understood with complex analysis -- see de Moivre's result which follows.)

Dynamics
A more intricate "exercise" is to imagine what happens if I take z=–1+i and start to move it. If I move it radially (slightly in or out) the corresponding fourth roots will vary also, of course. They will move in or out, always staying in a square centered at 0. What if I move z counterclockwise? In that case, the fourth roots will also move counterclockwise, but their rotational velocity will be one-fourth that of z. And if we move z fully around the origin, returning to 1+i, the vertices of the square will each be shifted cyclically, A to B, B to C, C to D, and D to A.

De Moivre's Theorem
This result is a special case of the reasoning we've just discussed. Here it is:

Who he? Abraham de Moivre was born in France (1667) and died in England (1754). Here is a neat short biography, which declares De Moivre is also remembered for his formula for (cos x + i sin x)n which took trigonometry into analysis, and was important in the early development of the theory of complex numbers. It appears in this form in a paper which de Moivre published in 1722, but a closely related formula had appeared in an earlier paper which de Moivre published in 1707. De Moivre studied Newton's Principia thoroughly. Another reference states Eventually de Moivre become so knowledgeable about the material that Newton would refer questions to him saying, "Go to Mr. de Moivre; he knows these things better than I do." That is quite a recommendation! De Moivre lived in poverty for most of his adult life. In the first part of his life, he experienced religious prejudice in France and then later, prejudice against immigrants in England. Human beings are both wonderful and consistent, and are sometimes not wonderful and still consistent.

Here are some problems which the techniques of this course can investigate or solve or compute efficiently.

1. The function arctan(x) has a graph which is very nice, a smooth strictly increasing curve through the origin. Also, we know from calc 2 that the Taylor series of arctan centered at the origin is ∑_n=0^∞(–1)ⁿx²ⁿ⁺¹/(2n+1). The radius of convergence of this series is 1. What the heck goes wrong when x→1^– (or maybe x→–1⁺)? Is there some invisible flaw in the function or tiny "kink" in the graph which causes a problem? What is happening?
2. There are many techniques for solving differential equations, and we have several courses devoted to these techniques. One method is to transform the equation using the Laplace transform or the Fourier transform, and then analyze the result. Part of the resulting computational "baggage" is computing specific definite integrals. For example, the following improper definite integral,
   ∫_–∞^∞[cos(mx)/(1+x²)]dx
   (here m is an arbitrary real parameter) comes up when computing Fourier transforms. The definite integral converges because the top is between –1 and +1, and the bottom, which is never 0, decays like 1/|x|² when |x|→∞. In practice the exact value of this integral is needed. Most standard computer algebra systems (surely Maple or Mathematica) can give a (remarkably!) simple answer. How do they do it? If you think about some of the standard "tricks" to compute integrals (integration by parts? substitution?) you would probably realize that they are inappropriate in this case. How do these programs compute the integral? How difficult is it and can this computation be done "by hand"?
3. One of the very classical models of many physical situations involves the following partial differential equation: Δh=∂²h/∂x²+∂²h/∂y²=0. One phenomenon (among many) which this models is "steady state heat flow in an isotropic medium": that is, say, a thin metal plate where a boundary temperature is prescribed and the temperature inside the plate doesn't change with time. A solution of the differential equation is called a harmonic function. The combination of the differential equation (Laplace's equation) and the boundary condition is called the Dirichlet problem. It turns out that, under fairly mild assumptions, there is a unique solution to the Dirichlet problem. Maybe this is not obvious, but the way to convince yourself that this is true is to start heating or cooling the edge of a plate, and just let the heat "propagate" until an equilibrium is reached. The same partial differential equation also models other phenomena. O.k.: there is a solution. But how can we actually find or compute the solution? For example, we could "prescribe" the boundary temperature on the unit disc. One (seeming!) very simple boundary distribution might be temperature=1 on the upper half of the boundary, and temperature=0 on the lower half of the boundary. If then we want an equilibrium temperature distribution on the inside of a disc-shaped plate, my guess is that the inside temperature will be between 0 and 1 (hey, if a point inside had temperature 150, I bet it would leak lots of heat to the boundary). I also think that the symmetry of the boundary temperature would result in symmetric isothermal curves inside the disc. O.k.: I can do this much by "pure thought", but how can I now actually compute the inside temperature? What are those curves?

Just as in real calculus the variable is usually called x, in complex analysis we call "the" variable is usually called z. And, just to quiet speculation, the following equation is true:

 d
---(z5)=5z4
 dz

Introducing the main characters
The plane is a wonderful geometric object, and it can be thought about in many different ways. Let me consider three of them.

• Two dimensional vectors: forces or velocities or ... So these are arrows with a head and a tail. They are directed line segments and the length is the magnitude, etc. They can be added by putting head to tail, etc. So then every vector can be written as a sum of i and j and ...
• R²: by this I mean the collection of all ordered pairs (a,b), with a and b real numbers. We meet this formulation (hey: the two-dimensional real vector space: formidable language!) for example in linear algebra, where we learn to manipulate vectors (actually in any dimension) using row reduction. We can multiply vectors by scalars, add them, etc.
• The plane is also C, the complex numbers. Here I think of a+bi (with a and b real) as a number, and I will assume that i²=–1. Then I will try to follow my nose (?) and compute with these things.

What's a field?
To me, all of linear algebra is one algorithm plus lots of definitions. Of course, this is a simplification, but not a huge one. The algorithm has various names: I usually think of it as row reduction. The definitions help us understand how to cope with the information going in and out of the algorithm. To "run" row reduction, we need to be able to add and multiply and to subtract and divide and ... Well, here are the organizing principles which seem to be necessary.

Addition Addition is commutative and associative; there is an additive identity and there are additive inverses.

Commutative means that the order of addition doesn't matter: 3+7=7+3. Associative means that the grouping doesn't matter: 3+(7+4)=(3+7)+4. The additive identity means there is a number 0 so that 7+0=7 (for "all values of 7"). And of course additive inverse means that there is a number –7 so that 7+(–7)=0.

The words have meanings similar to what they mean for addition. The only interesting variation is that 0 can't have a multiplicative inverse ("You can't divide by 0.") because people learned this was necessary to avoid some serious logical and computational problems.

A very tiny field
0 (the additive identity) and 1 (the multiplicative identity) had better be different. But just with those two numbers there is enough to do arithmetic. What follows are tables defining addition and multiplication. Certainly a century ago this was merely a mathematical curiosity.

  + | 0 | 1      + | 0 | 1
 -----------    -----------
  0 | 0 | 1      0 | 0 | 0
 -----------    -----------
  1 | 1 | 0      1 | 0 | 1 

The field we grew up with
I want to count: 1, 2, 3, 4, 5, .... and divide up pies (half a pie or a quarter of a pie for me, and a tenth of a pie for you). So we are led to the rational numbers, Q, a very natural field. I like this field. With patience, I can even compute moderately efficiently with it. However, there is something wrong with it: a lack of completeness which is upsetting to our intuition.
Very frequently people are treated to the proof that the square root of 2 is irrational. Well, I think the proof really says the following: there is no rational number whose square is 2. Maybe that's not surprising. But let's recast it. Consider the graph of y=x²–2 on the rational plane. That is, we only look at points whose coordinates are rational. There are lots and lots of rationals, so I bet that the graph looks like what is drawn to the right (heck, the picture and web browser actually use finite field algorithms, not even rational algorithms!). Since no rational number has square equal to 2, the curve can't intersect the horizontal axis. But look: the graph has points above the axis and points below the axis. So it magically (?) goes from below to above without hitting the axis. This is the upsetting part.

Completing that one
If you want to have enough numbers so that things intersect correctly you need the real field. That's the one with all possible decimal expansions. We have a course (Math 311) mostly devoted to proving everything you'd want to know about the real numbers using this completeness assumption. So I don't need to do anything about it. But, sigh, there is something even simpler that's still wrong. In high school we learn that polynomials should have roots. So the embarrassment of x²–2 is dealt with by assuming completeness (the parabola and the line should intersect!). But a more obvious flaw is that x²+1 which should clearly (?) have two roots doesn't have any at all!

Correcting the flaw
So the complex numbers, C, correct the flaw. In C, x²+1 (or z²+1, to use the variable name which is more common) has two roots, +/–i.

Arithmetic in C
I want to be sure that we can do arithmetic in C. So let me write how to add and multiply complex numbers.
Suppose z=a+bi and w=c+di. Then:
Addition z+w=(a+c)+(b+d)i. (This is exactly the old "head-to-tail" definition of vector addition. Good: less need to enlarge the brain.)
Multiplication Well, if we just try to compute (a+bi)(c+di) and assume everything will work (distribute, use i²=–1, etc.), then zw=(ac–bd)+(ad+bc)i. So that's our definition.

Verification of the field rules in C (eh)
Clearly these rules make C into a field. Well, maybe not clearly. But I am lazy and verification that C is a field is tedious. There is one aspect that takes some thinking, so let me address it, and that's the existence of multiplicative inverses. I want to show you that if a+bi is not the 0 complex number (0+0i) then there is another complex number whose product with this one gives me 1 (1+0i). Here is a way to think about it:

  1       1      (a–bi)        a–bi        a–bi  
----- = ----- · -------- = ------------ = ------
 a+bi    a+bi    (a–bi)    (a+bi)(a–bi)    a2+b2

Indeed, a(a/[a²+b²])–b·{–(b/[a²+b²])}=[a²+b²]/[a²+b²]=1 (that's the first part of the product) and a(–b/[a²+b²])+b·(a/[a²+b²])=[–ab+ba]/[a²+b²]=0 (that's the second part of the product).
Notice, please, that the above computation is successful exactly when the number a²+b² is not 0, and this is exactly when the complex number a+bi is not the 0 complex number, 0+0i.
A sane person might ask what is going on here. Well, my primary answer would be, "It works, doesn't it?" If that is not satisfactory to you, then there are a large number of detours I could suggest (study some algebra -- for example, Math 451-452). But for this course, I will try to make believability arguments (heuristic discussions). This is actually how things grew historically.

Vocabulary and pictures
Almost everything that "accidentally" got written above turns out to have a special name. So if z=a+bi:

• a is called the real part of z, and we write a=Re(z).
• b is called the imaginary part of z, and we write b=Im(z).
• The number sqrt(a²+b²) is written |z|, and is called (variously) the modulus of z, the norm of z, the magnitude of z ... I'll probably just use the first word most of the time. The reason to think about the square root instead of what's inside is because of the picture to the right.
  Note Square roots are at the core of a great deal of difficulty in this subject. Right now I will only use sqrt to mean the following: if V is a non-negative real number then sqrt(V) will mean the unique non-negative real number which, when multiplied by itself (its square) gives V. I'll work more diligently to discuss how we should think about other square roots.
• a–bi is called the complex conjugate of z, and is written z with a bar over it (in html I will write z) and we sometimes say z-bar. Geometrically, z is the reflection of z in the horizontal axis.

Relabeling the pictures
Let's draw things a bit differently and get a polar representation of z. Then we get the interesting formula
   z=r(cos(θ)+i sin(θ)) (somehow I usually write the i in front of the sine)
Here r is the modulus, and θ is called the argument of z. And now we are in a great deal of trouble, because of "the". Actually, there are many arguments which are valid. We only need the information contained in the values of sine and cosine of θ, so arguments are really only defined up to a multiple of 2Π.

The difficulty is actually fundamental. Let me show you: let's start with a point on the positive real axis, A. I guess I could have its argument be 0. Then let us travel around towards B. If I want the argument to move or vary continuously, then I think that B's argument should be Π/2. Now let's go to C on the path shown. C is on the negative real axis, and, to keep things nice, it should have argument Π. Of course we will move on to D, which should have argument 3Π/2. Now creep very slowly to E, and E itself is darn close to A. Well, I bet that E's argument, if we want things to vary continuously, should be nearly 2Π. But ... but ... if we let E move to coincide with A, then what should the argument be? Is it 0 or 2Π? Have we somehow made a mistake? Not at all: this is part of what we need to explain. If we want to have a continuous argument, then we will have to agree that argument is defined only up to integer multiples of 2Π (as math folks might say, "mod 2Π"). This will come up again below in a more algebraic context. This is a serious complication which won't go away.

A wonderful computation
Complex number addition is just a rephrasing of the old (2-dimensional) vector addition. Complex multiplication (with ac–bd and ad+bc) seems more complicated. Let me show you a neat computation that might help clarify what is going on.

Suppose z=r(cos(θ)+i sin(θ)) and w=s(cos(λ)+i sin(λ)). Let's compute zw. We can do this in any order (associativity) so that
zw=rs(cos(θ)+i sin(θ)) (cos(λ)+i sin(λ))=rs(cos(θ)cos(λ)–sin(θ)sin(λ)+i cos(θ)sin(λ)+sin(θ)cos(λ)).
We accidentally recognize certain things. (In fact, there aren't any accidents. Later you will see why these computations work as they do.).
zw=rs(cos(θ+λ)+i sin(θ+λ)). Well, cos(thing)+i sin(thing) always has modulus 1 (because sin²+cos²=1). So:

The consequences
The modulus of the product is the product of the moduli: |zw|=|z||w|.
The argument of the product is the sum of the arguments.

These statements help understand what complex multiplication "does" because the formulas ac–bd and ad+bc seem complicated. Consider the function mapping z→iz (that is, multiplication by i). Then the modulus multiplies, so |iz|=|i| |z|=1 |z|=|z| since i has modulus equal to 1. Therefore iz must be a complex number on the circle centered at 0 with radius equal to |z|. What about the argument? Well, arg(iz)=arg(i)+arg(z)=Π/2+arg(z), so the argument of iz is a right-angle plus the argument of z. The plus is important since it tells us that iz is z rotated clockwise by a right-angle (relative to the origin). So the mapping z→iz is a rotation by a right angle around the origin.

But all this has some difficulties (I'd like to be sort of honest!). I know that:
i²=–1, and i³=–i, and i⁴=1.
If we take arg(i) to be Π/2, then arg(i²)=2(Π/2)=Π and arg(i³)=3(Π/2)=3Π/2 and arg(i⁴)=4(Π/2)=2Π, which might contrast interestingly with a naive value of arg(i⁴)=arg(1)=0. So 0=2Π? Well, "mod 2Π" or, up to an integer multiple of 2Π, that's true. We need to keep this in mind. It will come up again and again.

Maintained by greenfie@math.rutgers.edu and last modified 1/20/2010.