Isolated singularity: the definition
f(z) has an isolated singularity at z₀ if there is r>0 so that f(z) is analytic in the "deleted disc", 0<|z-z₀|<r.
Comments This definiton leads to some silly examples. Let's see: the function f(z)=z² has an isolated singularity at, let's see, -4=5i or, in fact, any z₀. That's a consequence of the language. The examples we "really" want to cover imply that we need this rather weak definition.

Isolated singularity: here's all you need to know
Here's a table with all the entries completed. Please realize that Inventing" and examining the entries required about 1.5 class meetings -- quite a lot of time. The most amzing fact is perhaps hidden in the trichotomous structure of the table itself: any isolated singularity must be one of the named varieties. There is no other kind of allowed behavior. This is in sharp contrast with the free situation in one-variable (real) calculus. There if you know that a function is, say, differentiable in all of R except 0, all sorts of weirdness can occur.

Isolated Singularities
Name	Characteristic example(s)	Defining behavior	Relevant theorems	Characterization by Laurent series
Removable singularity	[sin(z)]/z at z0=0	There's a way of defining the function, f(z), at the center z0 of the deleted disc, so that when the f(z) is looked at with this value, the reuslt is analytic in all of the disc, |z-z0|<r	Riemann's Removable Singularity Theorem If f(z) is bounded in 0<|z-z0|<r, then f(z) has a removable singularity at z0. "Bounded" means there is a real (positive) M so that |f(z)|<M for all z's with 0<|z-z0|<r	All of the coefficients in the Laurent series for f(z) in the deleted disc centered at z0 corresponding to negative integer indices are 0. The Laurent series is a power series.
Pole	7/(z-5)3 1/(ez-1)2 at z0=0.	The limit of |f(z)| as z-->z0 is infinity. That is, given any positive real number M, there is some epsilon>0 so that if |z-z0|<epsilon, |f(z)|>M.	In the deleted disc, f(z)=h(z)/[(z-z0)k, where h(z) is analytic in the whole disc |z-z0|<r and and h(z0) is not 0. k is a positive integer called the order of the pole at z0.	Only a finite number of terms in the Laurent series for f(z) in the deleted disc centered at z0 corresponding to negative integers are not 0. You can "factor out" the correct inverse power of z-z0 from the Laurent series and get [(z-z0)k mulitplying a power series with a non-zero constant term.
Essential singularity	e1/z	f(z) is neither a pole nor a removable singularity.	Casorati-Weierstrass Theorem In any deleted disc (no matter how small!) centered at z0, you can find arbitrarily close approximate solutions to any equation of the form f(z)=w0. More precisely, given a>0 and b>0 and any w0 in C, there exists z in 0<|z-z0|<a so that |f(z)-w0|<b.	There are an infinite number of non-zero Laurent coefficients with negative integer indices. (Hey, this says nothing about the coefficients with positive indices!)

Example of a ludicrous result
Suppose that f(z) has an isolated singularity at 0. Suppose also that we know there are two sequences, {z_n} and {w_n}, of complex numbers which both converge to 0. Suppose even further that the sequences {f(z_n)} and {f(w_n)} both converge, and that the first sequence has limit 7 and the second sequence has limit 8.
Then there is a sequence of complex numbers, {q_n} which converges to 0 so that {f(q_n)} converges to 1+i. Proof Well, since there is a sequence approaching 0 whose function values have a finite limit, the isolated ingularity can't be a pole. If the isolated singularity were removable, then any sequence approaching 0 would have function values approaching f(0), so there would be exactly one limit of the sequence of function values. But we're told that there are two distinct limits (7 and 8). Therefore, the singularity is essential. So now take a=1/n and b=1/n and w₀=1+i in the Casorati-Weierstrass Theorem. Let me call the "z" produced by that theorem, q_n. Well then, the sequence {q_n} has |q_n|<1/n so the sequence converges to 0. Also, |f(q_n)-(1+i|<1/n, so the sequence {f(q_n} converges to 1+i.

The uselessness of mathematics (Is this news?
I don't have enough time or space to go into detial on this topic, so I'll just discuss the relevance of "uselessness" to the previous theorem. The theorem is what's called an existence result. There is a sequence of q_n's (in fact, there are many, many such sequences. The darn theorem doesn't tell you anything about the sequence. It offers no procedure, no algorithm, for approximation. In some sense, the theorem is totally useless. And any "random" essential singluarity is indeed very difficult to analyze computationally. It could be very difficult to find such a sequence, even approximately.

Analyticity in an annulus yields Laurent series
An annulus in C is a region defined by two real numbers, a<b (here a>=0 and b<=infinity) and a center, z₀. The annulus resulting is the collection of z's satisfying a&lot;|z-z₀}<b. One extreme example can be obtained by taking z₀=0 and a=0 and b=infinity. Then the annular region is all of C except 0. It turns out that if f(z) is analytic in an annular region, then f(z) is equal to the sum of a certain series, a series which is sort of like a power series but which can have terms corresponding to all integer powers of z-z₀, not just positive integer powers. So for example, if I wanted to look at, say, the function f(z)=e^5z+e^([2+8i]/z), I bet that I can replace this function by
SUM_n=0^infinity[5ⁿ/n!]zⁿSUM_n=0^infinity[(2+8iⁿ/n!]z^-n.
This is really two series, of course, and both of them converge separately. The whole thing is a way of writing f(z) as the sum of powers of z^integer, where the integer can be positive or zero or negative. Of course, there are limits involved in the definition of the infinite series (for both, I need the limit of the partial sums to converge) but it turns out that, given any function f(z) analytic in an annulus, there is always exactly one such series (a doubly infinite series in powers of z-z₀) whose sum is equal to f(z).

The exposition in the textbook is good (pp. 141-144), and shows how, given a function f(z) which is analytic in an annulus, then there are unique complex constants a_n, for any integer n, so that f(z) is equal to SUM_n=-infinity^+infinitya_n(z-z₀)ⁿ. Here I think the simplest way to understand the doubly infinite sum is to think of it as the sum to two infinite series. What's really neat is that this series converges in a manner similar to power series inside the radius of convergence. That is, it converges absolutely, so you can do anything algebraic you would like (rearrangement, grouping, etc.). Also, the series converges uniformly on circles |z-z₀|=t where t is between the inner and outer radius of the annulus. This means you can "exchange" integral and sum, and not worry. These properties follow in a similar fashion as they did for power series. One "expands the Cauchy kernel" and the geometric series plays a double role, on an inside and an outside circle. I gave a very rapid, not so good, exposition of the "expanding" process. I think I got the formula a_n=_{|z-z0|=tf(z)/(z-z}0)ⁿ⁺¹ dz.

I then applied the properties of Laurent series to get the characterizations for an isolated singularity. If the isolated singularity is removable, then consider the formula for the a_n's above when n<0. Look at the signs of the power inside the integral, and see that for negative n's, the integrand is analytic inside of |z-z₀|<r for a removable singularity.

If f(z) has a pole at z₀, then 1/f(z) has a zero there (and the singularity is removable). Look at the power series for 1/f(z). You can "factor out" the lowest power of z-z₀ which appears, and the result is (z-z₀)^k multiplying a power series whose constant term is not 0. But the power series represents an analytic function p(z) with p(z₀) not equal to 0. Therefore f(z)=[1/(z-z₀)]h(z) where h(z)=1/p(z) is analytic near z₀ and which therefore has a power series beginning with a non-zero constant term. Wow. Then multiply through and get the Laurent series behavior guaranteed by the table. The Laurent characterization of essential singularities follows from the fact(s) that the other Laurent characterizations are "if and only if" -- no negative terms if and only if removable; finitely many negative terms if and only if pole.

One computation
I think here I asked a Rutgers qualifying exam question (spring 1999): find the order of the pole at 0 of 1/(6sin(z)+z³-6z)²?
Well, I know that sin(z)=z-z³/3!+z⁵/5!+... so that 6sin(z)+z³-6z is z⁵/20+higher order terms. Thus, 6sin(z)+z³-6z=z⁵(a power series whose constant term is 1/20)=z⁵k(z), where k(z) is an analytic function and k(0)=1/20. Then this in turn means that 1/(6sin(z)+z³-6z)² is z^-10h(z) where h(z) is analytic and h(0)=400 (all I care about here is that h(0) is not 0). Therefore the order of the power is 10.

I found (some of_ the Laurent series for each of the functions in the table.

The child's residue theorem
Suppose f(z) has an isolated singularity at z₀, so f(z) is analytic in 0<|z-z₀|<r. If 0<t<r, then _|z-z0|=tf(z)=2Piia_-1 where a_-1 is the -1^th coefficient in the Laurent expansion of f(z) in 0<|z-z₀|<r.
Proof Well, interchange integral and Laurent series sum. we just need to compute _|z-z0|=t(z-z₀)ⁿdz. If n>=0, the integrand (the function we are integrating) is analytic, so the integral is 0 (Cauchy's Theorem). Now if n<-1, the integrand (z-z₀)ⁿ has an antiderivative in the deleted disc. This is, of course, [1/(n+1)](z-z₀)ⁿ⁺¹. But our version of the Fundamental Theorem of Calculus says that functions which have antiderivatives (primitive) must have integrals equal to 0 over closed curves. So the only integral which could be non-zero is _|z-z0|=t(z-z₀)^-1dz and this is non-zero for any of many reasons. Certainly we could (we have!) directly computed it very early in the course, and it has value 2Pii.

The residue of f(z) at z₀ is a_-1, the (-1)^th coefficient of the Laurent series of f(z) centered at z₀

HOMEWORK
2.4: 20, 24a
2.5: 1, 6, 7, 12, 14, 22b

Last time we learned "another result making analytic functions equation. We can apply this to
An exam problem
This one is from a written qualifying exam in mathematics given by the University of California.
Suppose f(z) is analytic and for all positive integers k, f({sqrt(-1)}/k}=100/k⁴. What is f(z) (and why)?

An answer
Well, I hope we can guess an answer. How about f(z)=100z⁴. Direct evaluation shows that this function does fulfill the given equation(s). The most important observation is that the fourth power of i is 1! Are there any other functions? The answer is "No," because if there were, then another function would be equal to 100z⁴ on a convergent sequence, and then would have to be indentical with this.

Radchenko-Bolzano-Weierstrass
I discussed Mr. Radchendo's observation. I had another motive because I needed the Bolzano-Weierstrass Theorem below. Heh, heh, heh.

The Cauchy estimates
These are the contents of problem #18 of section 2.4. They follow directly from the Cauchy integral formula for derivatives. Use the ML estimate to get |f⁽ⁿ⁾(z₀)|<=[n!/(2Pi)]LM where L is 2Pi r (the length of the circle of radius r centered at z₀) and M is the max of |f(w)/(w-z₀)ⁿ⁺¹| for |w-z₀|=r. This gives exactly the conclusion of problem #18:
|f⁽ⁿ⁾(z₀)|<=[n!/rⁿ]max_|z-z0|=r|f(z)|.
The |(w-z₀)ⁿ⁺¹| gives an rⁿ⁺¹ and the 2Pi's cancel.

Another exam problem
Suppose someone comes up to you on the street and gives you an entire function, f(z), with the following silly property:
|f(z)|<=4(1+|z|)¹⁰.
What can you say about f(z)?

The first time you see one of these problems there is a great amount of strangeness. One tries to think of examples. For this estimate, there certainly are lots of examples, such as z and z² and ... But then one tries to think of non-polynomial examples, such as sin(z). But we eliminated sin(z), because while it has rather tame "growth" on the real axis, if z=iy, then |sin(z)|=|sinh(y)) which grows exponentially (like (1/2)e^y) and can't be boun ded by poloynomial growth.

In fact, if |f(z)|<=4(1+|z|)¹⁰ for all z, then f(z) must be a polynomial of degree 10 at most. Why? The easiest way to verify that a function is a polynomial is to differentiate it a bunch of times and see that the result is identially 0. A function that has derivative 0 must be a constant. A function whose second derivative is 0 must be linear. A function whose ... In this case, I will prove that the 11^th derivative of f(z) is 0. I will use the Cauchy estimates with n=11.

So |f⁽¹¹⁾(z₀)|<=[11!/r¹¹]max_|z-z0|=r|f(z)|.
But |f(z)|<=4(1+|z|)¹⁰, so if |z-z₀|=r, then |f(z)|<=4(1+|z-z₀+z₀|)¹⁰<=4(1+|z-z₀|+|z₀|)¹⁰<=4(1+r+|z₀|)¹⁰ if |z-z₀|=r.
Combine these to get |f⁽¹¹⁾(z₀)|<=[11!/r¹¹]4(1+r+|z₀|)¹⁰ which is true for all r>0. In particular, look at what happens as r-->infinity. The complicated stuff has 11 powers of r on the bottom and 10 powers of r on the top. Hey: that means |f⁽¹¹⁾(z₀)| is overestimated by something that -->0. Therefore |f⁽¹¹⁾(z₀)|=0 for all z₀'s. Therefore f(z) must be a polynomial of degree at most 11.

And another
I think this is from a Rutgers exam. Here the estimate given is something like |f(z)|<=sqrt(1+|z|) for all z, and the question again is: what can you conclude? Hey, here is a hint and here is the answer.

Liouville ==> Fundamental Theorem of Algebra
We already saw this proof technique in its simplest version, Liouville's Theorem. Let me show you what I think is the fastest proof of the Fundamental Theorem of Algebra. I did this is class for an example. Let me try again, with a slightly different exposition.

Suppose p(z)=z⁷+(5+i)z⁶-4z²+z-9. Then the main idea is to look at p(z) for |z| large. |p(z)|=|z|⁷|1+(5+i)z^-1-4z^-5+z^-6-9z^-7|. Now choose |z| large enough. Let me try |z|>100. Then
     |(5+i)z^-1|<=sqrt(6)/100
     |-4z^-5|<=4/(100)⁵
     |z^-6|<=1/(100)⁶
     |-9z^-7|<=9/(100)⁷
So now look at |1+(5+i)z^-1-4z^-5+z^-6-9z^-7|. Use the reverse triangle inequality to conclude that this mess has modulus at least
1-(sqrt(6)/100+4/(100)⁵+1/(100)⁶+9/(100)⁷) and this is greater than, for example, 1/2.

If |z|>100, |p(z)|>=|z|⁷(1/2). Therefore |p(z)| can't be 0 and in fact, 1/p(z) will have modulus at most 2/(100)⁷ when |z|>100.

The approach to proving the Fundamental Theorme of Algebra is this: if p(z) is never 0, look at f(z)=1/p(z). I bet that this is bounded, hence (Liouville) constant, contradiction because p(z) is certainly not constant.

The estimate we just verified:
     1/p(z) will have modulus at most
     2/(100)⁷ when |z|>100.
takes care of verifying boundedness on the outside of the disc of radius 100 centered at 0. How about the inside?

Well consider p(z) on the z's with |z|<=100. If I assume that p(z) is never 0 (or else we have a root!) then what can I say? Well, suppose that |p(z)| gets close to 0. In fact, what if there is a sequence {z_n} inside this disc with |p(z_n|<1/n. Then (Bolzano-Weierstrass) there is a convergent subsequence. That is, we can find a subsequence so that |p(z_nk|<1/n_k and the z_nk-->some number w. But then p(w) has to be 0 since the polynomial p is continuous and p(w) is the limit of the p(z_nk)'s. But we are assuming that p has no roots. Therefore |p(z)| must have a positive minimum inside |z|<=100. So f(z)=1/p(z) has a positive maximum inside |z|<=100. Therefore f(z) is eligible for Liouville's Theorem, and hence we get the contradiction mentioned above.

Outside the disc, use a simple underestimate, and inside the disc, use Bolzano-Weierstrass to convinced yourself that p(z), if it were to never be 0, would have a positive minimum in modulus.

Order of the zero of an analytic function
Suppose that f(z) is 0 at z₀. Look at this sequence:
f(z₀), f´(z₀), f´´(z₀), f⁽³⁾(z₀), f⁽⁴⁾(z₀), f⁽⁵⁾(z₀), ...
The sequence certainly begins with 0. Could it be all 0? Well, if it is all zero, then f(z) must be the zero function. Otherwise some first number in the sequence is not zero. The number of the derivative of f which isn't 0 is called the order of the zero of f(z) at z₀.

Example
I think I did something like this:
look at f(z)=(cos(z)-1+(1/2)z²)⁵. Certainly f(0)=0. What is the order of the zero of f(z) at 0?
We could start differentiating f(z) and plugging in 0 and see when the result is not 0. Or we could be slightly devious.

Since f(z)=(cos(z)-1+(1/2)z²)⁵ and we know the Taylor series for cos(z) centered at 0, we can write
f(z)=([1-(1/2)z²+(1/24)z⁴+HOT]-1+(1/2)z²)⁵ whjere HOT stands for "higher order terms".
Then f(z) must be ([(1/24)z⁴+HOT])⁵. Now I hope you can see easily that f(z) has a zero of order 20 at 0 and even that the 20^th derivative of f at 0 has value (20!)/(24)⁵ (the 20! comes from the formula for the coefficients of the Taylor series).

Too many ideas ... let us reconsider
I decided I should go backwards and go over some of the tougher or more obscure things I discussed last time. So that's what I did, at least for the first hour.

THIS  IS  NOT  COMPLEX  ANALYSIS  SO  YOU  CAN  DISREGARD  IT! The perils of integral(sum)=sum(integral) One thing I did last time was interchange an integral and a sum. I wanted to spend some time on this. Indeed, we actually counted the math symbols in the equation I wrote, and discovered that there were 41 of them on just one side, and that's too darn many to understand. I wanted to simplify the situation as much as possible to show you how bad things could be. So instead of a complex line integral, I looked at "standard" real integrals of continuous functions on intervals of R. The function I began with is graphed to the right. f1(x) is a piecewise-linear continuous function, and is not 0 only in the interval [0,3k]. Here k will be some positive real number I don't really care about. For x<0 and for x>3k, f1(x)=0. Then the graph "connects the dots" (0,0), (k,1), (2k,0), (2k+k/2,-2) and (3k,0) linearly. Certainly the integral of f1(x) from 0 (or anything less than 0) to 3k (or anything greater than 3k) is 0. The triangles cancel out. Now to get f2(x). First, I'll double f1(x): consider 2f1(x). Then let's squeeze it so it will vibrate twice as fast: 2f1(2x). Now finally let us shift it right by 2k. Thus, f2(x)=2f1(2[x-2k]). The pictures below illustrate what I'm doing. (except for the last picture which I just noticed needs to be moved right another k!). Notice please that the integral of f2(x) is again 0. Notice that in the sum f1(x)+f2(x), the lower triangle of f1 is canceled by the upper triangle of f2. The sum of the two functions is 0 away from (0,3.5k). Well, I will "construct" or indicate how fn+1 should be defined by a recursive formula:     fn+1(x)=2fn(2[x-k/2n-2]) I think this is the formula I want. fn+1 should be a function which has one positive triangle of height 2n-1 and width k/2n-2, and a negative triangle of height 2n and width k/2n-1. Its total integral is 0. What is the most interesting about this is the finite sum, SUMn=1mf_n(x) where m is very large. It takes some effort to see that there is lots and lots of cancellation. The total integral is 0, this sum function "sits" inside the interval [0,3k], and it has one "stable" positive fat triangle to the left, and one very thin, very long negative triangle on the right. The total integral of this is 1. Now what is limm-->infinitySUMn=1mf_n(x)? In fact, if you think very carefully about this, the limit is 0 except for the stable positive fat triangle. The integral of this stable positive fat triangle is k, which isn't 0. But what about this limit: limm-->infinity03kSUMn=1mf_n(x) dx. Each of these integrals is 0, as we've noted, so the limit exists and is 0. We now see: limm-->infinity03kSUMn=1mf_n(x) dx=0 and 03klimm-->infinitySUMn=1mf_n(x) dx=k which is not 0. So infinite sum and integral may not "commute"! If you would prefer arguments involving functions defined by formulas, and if you would like to learn more about the mistakes people made when they began studying Fourier series and power series, one nice reference at a level appropriate for students in this class is A Radical Approach to Real Analysis by David Bressoud.

Uniform convergence?
There are conditions which guarantee equality when integral and sum are exchanged. One such condition is called uniform convergence. I'll just outline what happens. Sometimes Math 311 covers this material, although I did not in my last instantiation of 311.

Suppose that we have a sequence of functions, {f_n(x)}, each continuous and defined on the interval [a,b]. Then we say that SUM_n=1^infinityf_n(x) converges uniformly to F(x) on [a,b] if, given any epsilon>0, there is an N so that if m>N, |SUM_n=1^mf_n(x)-F(x)|<epsilon for all x on [a,b]. This condition certainly does not hold in the example previously discussed. But then
|_a^bSUM_n=1^mf_n(x)-F(x) dx|<=_a^b|SUM_n=1^mf_n(x)-F(x)| dx<=(b-a)epsilon for all m>N. This implies that lim_{m-->infinitya}^bSUM_n=1^mf_n(x) dx=_a^bF(x) dx.
If you are clever, you can see that we're using a version of the ML inequality to prove this. So uniform convergence imples that So infinite sum and integral do "commute" if uniform convergence is also assumed.

And power series almost do that ...
We compared power series inside their radius of convergence to a fixed convergence geometric series. We were really proving that in any subdisc which is strictly contained inside the radius of convergence, the power series converges uniformly. So we can actually interchange sum and integral.

Return to 1/(1+x²)
I asserted last time that, given any x₀ in R, 1/(1+x²) had a power series expansion centered at x₀ with radius of convergence at least 1. This is true but is certainly not obvious. How can I produce the power series expansion for you? I could try to differentiate the function and create a Taylor series. I think Maple does this:

series(1/(1+x^2),x=5,5);
 1    5           37              30              719 
-- - ---*(x-5) + ----*(x-5)^2 - -----*(x-5)^3 + -------(x-5)^4 + O((x-5)^5)
26   338         8788           28561           2970344

I know that 1/(1+x²)=A/(x-i)+B/(x+i). Here we can guess that A=-i/2 and B=i/2 (or you can systematically solve the partial fraction equations). Now let me look at the first piece.

We know that (i/2)/(x-i)= (i/2)/([x-5]+[5-i]). If x-5 is small, we can use the geometric series, with r=(x-5)/(5-i). So look:
(i/2)/([x-5]+[5-i])={-(i/2)/(i-5)}{1/[1-(-r)]}={-(i/2)/(i-5)}SUM_n=0^infinity(-1)ⁿ[(x-5)/(5-i)]ⁿ=(i/2)SUM_n=0^infinity[(-1)ⁿ/(5-i)ⁿ⁺¹](x-5)ⁿ.
So we have a piece of the power series. We can do something analogous with the other piece, (i/2)/(x+i). We will get (i/2)SUM_n=0^infinity[(-1)ⁿ⁺¹/(5-i)ⁿ⁺¹](x-5)ⁿ, I think.

If you add up the two pieces, you'll get the total power series for 1/(1+x²) centered at x₀=5. What's the radius of convergence of this series? We need r=(x-5)/(5-i) to have modulus less than 1. When does this occur? The modulus of 5-i is sqrt{26}. So actually the radius of convergence is sqrt{26}, and why is this?

Well, 1/(1+x²) is just the restiction to R of 1/(1+z²) analytic on all of C everywhere except +/-i. If you look at the proof we gave last time for the implication {analytic==>power series} you should see that we can get a power series for an analytic function up to the first "singularity", the largest disc so that the function stays analytic in the whole disc. Here, for 1/(1+z²) and for the center 5, that disc will have radius sqrt{26}, the distance from 5 to +/-i. In general, the radius of convergence of the power series for 1/(1+x²) centered at x₀ in R will be sqrt{1+x₀²} for the same reason. Therefore this function is a counterexample to the real version of the bubbles result I stated last time. The function is equal to a power series centered at every point, and the radius of convergence is at least 1. But the function is not equal to a power series with infinite radius of convergence.

Real/complex: what is the correct idea?
Complex results usually turn out to be much simpler. Therefore if you can turn a real problem into a complex problem, it is likely that the problem will be simpler. Probably. Maybe. Anyway, that's what many people think.

A refreshing break?
We took a ten-minute break. Snacks were served.

What is (or should be) sin(z)?
The textbook and class lectures have given two or three ways of defining a function which should be sin(z). That is, we would like an entire function, that is, a function analytic on all of C, which when restricted to R is the standard sine function. How many candidates are there for such a function, and how would we identify such candidates.

Derivatives at 0?
Well, if I look at the standard sin(x) function defined in calculus, then I know that sin(0)=0, sin´(0)=1, sin´´(0)=0, sin´´´(0)=-1, and then the values begin to cycle. If there is an entire function which should be sine, then, as we saw last time, that function should have a power series centered at 0 (actually the series could be centered at any point!) whose radius of convergence is infinity. But now plug in the numbers, and get the complex power series SUM_n=1^infinity[(-1)ⁿ/(2n+1)!]z²ⁿ⁺¹. This is the only power series centered at 0 which has the correct derivatives. Therefore the function defined by this series (whose radius of convergence is infinity!) must be sine, and is sin(z).

One result making analytic functions equal
We can generalize this result.

IF f(z) and g(z) are analytic in a domain D, and if there is a point z0 in D so that the values of f and g and all their derivatives agree at z0 THEN f(z)=g(z) in all of D.

A proof of essentially this result is on p.127 of the text. How does the proof go? First, because the function values and all the derivatives agree at z₀, the power series at z₀ for f(z) and g(z), which are Taylor series for z₀, must be the same. Therefore, the functions f(z) and g(z), which are equal in a disc to the sum of these series, must agree. But then we want to extend the agreement to all of the domain. Remember that domains are connected open sets. So connect z₀ with any other point by a curve inside the domain. Use that curve to draw a chain of discs each of which has its center inside the disc before it. Then inductively, f(z)=g(z) on one disc, so all the derivatives are equal so that the power series for each function at the next center will be equal, making f(z)=g(z) on that disc, etc.

Analytic functions as infinite degree polynomials
Sometimes I think that analytic functions are just infinite degree polynomials. And sometimes the analytic functions agree with me (?) and behave well. One way we could get two polynomials to be equal everywhere is to get them to agree at sufficiently many points. For example, if P(x) and Q(x) are polynomials on R, of degree, say, at most 14, and if the set where P(x)=Q(x) has at least 15 points, then P(x) and Q(x) must be the same polynomials everywhere. This sort of implication is used everywhere in numerical analysis, but is probably easiest to see using linear algebra. Notice that P(x) and Q(x) have 15 parameters (their coefficients) and making them agree at 15 points gives 15 linear equations in these parameters. The coefficient matrix involved in this system of linear equations is a Vandermonde matrix which has a non-zero determinant.

With this as background, maybe we can "force" two analytic functions to be equal by asking that their values agree at infinitely many points. Well, things aren't that simple. As suggested by several students, the entire functions sin(z) and cos(z) agree at infinitely many points (say, many [not all!] multiples of Pi/4) and certainly are not the same. But something like this almost works if you ask for a little bit more.

How to force a function to be sine
Suppose I know that f(z) is an entire function and, for example, I also know that f(1/2^k) is equal to sin(1/2^k). Then, in fact, f(z) must be sin(z), and, certainly, for all real x, f(x)=sin(x).
This is a remarkable result.
Why is it true? Well, since sin(x) is continuous, sin(0)=lim_k-->infinitysin(1/2^k). But this is the same, since f(z) is continuous and the numbers inside the limit agree, as f(0)=lim_k-->infinitysin(1/2^k).
But ... but ... maybe it would be better to prove this in general.

Another result making analytic functions equal

IF f(z) and g(z) are analytic in a domain D, and there is a sequence {zk} with f(zk)=g(zk) for all k and, finally, limk-->infinityzk=z0, a point in D, THEN f(z)=g(z) in all of D.

To verify this, we just need to show that f(z₀)=g(z₀), f´(z₀)=g´(z₀), f´´(z₀)=g´´(z₀), f´´´(z₀)=g´´´(z₀), etc. because then we can use the previous result.

Why is f(z₀)=g(z₀)?
Both f and g are analytic so they are continuous. Therefore since we know that lim_k-->infinityz_k=z₀, we see lim_k-->infinityf(z_k)=f(z₀), and lim_k-->infinityg(z_k)=g(z₀). But f(z_k)=g(z_k) for all k, so f(z₀)=g(z₀).

Why is f´(z₀)=g´(z₀)?
Look at the power series for f centered at z₀: f(z)=SUM_n=0^infinitya_n(z-z₀)ⁿ. And look at the power series for g centered at z₀: g(z)=SUM_n=0^infinitya_n(z-z₀)ⁿ. Remember that each series is a Taylor series. Since f(z₀)=g(z₀), a₀=b₀. Now consider the functions ff(z)=[{f(z)-a₀}/(z-z₀)] and gg(z)=[{g(z)-b₀}/(z-z₀)]. Although we seem to be dividing by z-z₀ and therefore making trouble, in fact, the series for f(z)-a₀ and the series for g(z)-b₀ both have common factors of z-z₀, so we are just canceling just common factors, non-zero away from z₀. The results are convergent power series and therefore are analytic functions. The functions ff and gg agree at each of the complex numbers z_k. Therefore, just as in the previous argument, ff(z₀)=gg(z₀), and that means a₁=b₁.

Why is f´´(z₀)=g´´(z₀)
Let us write two new functions: fff(z)=[{f(z)-a₀-a₁(z-z₀)}/(z-z₀)²] and ggg(z)=[{g(z)-b₀-b₁(z-z₀)}/(z-z₀)²]. These functions are also analytic, because although we are dividing by (z-z₀)² we have taken away the only parts of the power series where this would be an obstable. Also I know that fff(z_k)=ggg(z_k) since f(z_k)=g(z_k) and a₀=b₀ and a₁=b₁. Whew. Just as twice before, we now conclude that fff(z₀)=ggg(z₀), and that means a₂=b₂.

Why is ....
Well, why is the instructor so stubborn that he won't give a proof using mathematical induction? Well, it will work and we will get all of the coefficients of the power series for f and g centered at z₀ to agree, and therefore f(z)=g(z) in all of the domain.

So what is sine?
Well look at f(z)=sin(x)cosh(y)+icos(x)sinh(y). This function is analytic because it satisfies the Cauchy-Riemann equations. But notice that since cosh(0)=1 and sinh(y)=0, f(x+i0)=sin(x) for all real x. Hey, that certainly satisfies the hypotheses of the result we've just proved. So f(z) must be sin(z) for all complex z!

How to prove (?!) a trig identity
This result was proved with the help of Cohen²: why does (sin(z))²+(cos(z))²=1? Both sides of this equation are entire functions. The equality is true on all of the real numbers. Therefore, since the real numbers have a convergent sequence (!!), the equality is true for all of C.

Analytic continuation
Both of the results we've proved are at the beginning of what's called analytic continuation. We take functions and equations that work on a small set, and then, "automatically", things will work on a larger set. This is weird, remarkable, and useful.

A corollary observed by Mr. Radchenko
I'll mention this next time, but look: Mr. Radchenko declared that although having f(z)=g(z) at an infinite number of z's isn't enough to guarantee that f(z) must equal g(z) everywhere (cosine and sine!), if we ask that f(z)=g(z) at an infinite number of z's which are bounded, then they must be equal.
Why is this? Because (and this is a result from advanced calculus):

Bolzano-Weierstrass Theorem Suppose (xn) is a sequence all of whose terms are in [a,b]. Then (xn) has a convergent subsequence.

There's a proof using "bisection" in my 311 diary for 3/12/2003. Sigh. Look at 311, please.
If you are willing to believe this result, then you get a sequence where f(z)=g(z) which has a limit point, and ... you're done!

HOMEWORK
Read the material of 2.3 and 2.4. I will spend a considerable amount of time discussing this material next week. This is certainly a big chunk of the core of the course, the "standard" applications of the big theorems. We will need considerable course time to absorb this material.
Please hand in these problems on Thursday, March 31:
2.3 5, 18 (We proved 17 in class so please use it!).
2.4: 1, 3, 5, 13.

1. Power series
   We saw that the sum of a power series was analytic, and that the derivative of this sum is equal to the sum of the terms of the series differentiated: that is, a convergent power series behaves very much like we'd believe an "infinite" polynomial to behave.
2. Cauchy's Theorem
   We started with an analytic function defined in a simply connected domain. (One way of recognizing such: "No holes.") Then the integral of the analytic function around any simple closed curve is 0. I proved this with Green's Theorem, and I needed to assume that the analytic function had a continuous derivative. That turns out not to be necessary ("Goursat") but right now I don't care!
3. Integrals around closed rectangular paths are 0
   Suppose that a path in a simply connected domain is closed and is made up of a finite number of horizontal and vertical line segments. Then the integral of an analytic function around this path is 0.
4. Primitives of certain functions ...
   Suppose we have a continuous function defined in a domain, and we know that the integral of this function around a rectangle whose interior is in the domain is 0. Then the function has a primitive: that is, the function is the derivative of an analytic function. (Here we "constructed" the primitive by a process resembling what's done in physics and third semester calculus to get a potential from a conservative vector field.)
5. Cauchy Integral Formula
   This result, which I will abbreviate as CIF, is really weird. If you have a function f(z) that's analytic on and inside a simple closed curve, C, and if z is inside the curve, then f(z)=[1/(2Pi i)]_C[f(w)/(w-z)]dw.

Expanding the Cauchy kernel
This title refers to a really remarkable process which, even though it is 170 years old, is still interesting to me. We use it to get much information from the CIF. The Cauchy kernel is 1/(w-z). Here is the setup:
f(z) is analytic in a domain D. We select z₀ in D. We select a positive radius r so that the circle of radius r centered at z₀ and its inside are contained entirely within D. Also take any z inside the circle. Then the CIF implies that
      f(z)=[1/(2Pi i)]_|w-z0|=r[f(w)/(w-z)]dw.
Let's look at 1/(w-z), where w is on the circle. Then, as I showed in class, and is clearly written in the text, we write:
1/(w-z)=1/(w   -z)=1/(w - + -z)=1/(w-z₀+z₀-z).
This sort of thing is all routine now, but please try to imagine the excitement of inventing (discovering?) it. Now also realize that for w on the circle, |w-z₀|=r and since z is insided the circle, |w-z|<r. Now we need other neurons to "fire" and some synapses to connect. Well, |(w-z)/(w-z₀|<1. And so
1/(w-z₀+z₀-z)=[1/(w-z₀)][1/(1-s)] where s=[(w-z)/(w-z₀)] is a complex number whose modulus is less than 1. So we can use the geometric series, which I have mentioned again and again is almost the only series we really need to know.

Interchanging sum and integral
I interchanged the sum (which resulted from e-x-p-a-n-d-i-n-g the Cauchy kernel) and the integral. I remarked that this was a mathematically perilous step and I will go into this more in the next class. In this case, with a geometric series, we can make very explicit error estimates, comparing partial sums and the various infinite tails of the series. This is done in the text and students should at least look at the material.

After you do the exchange of sum and integral, and then "recognize" what you have logically, you will realize that:
f(z)=SUM_n=0^infinitya_n(z-z₀)ⁿ, where the a_n's are coefficients and magically given by the formula a_n=[1/(2Pi i)]_|w-z0|=r[f(w)/(w-z)ⁿ⁺¹]dw.

Power series forever!
Apparently we just proved that an analytic function must have a power series expansion which is valid in any disc contained in its domain. Let's see: since we can differentiate power series any number of times. Therefore we have proved:
The derivative of an analytic function is analytic.
So one derivative implies many, many more! Certainly in calculus this is false, since there are many functions which give non-differentiable results when you differentiate them.
Also please realize that we already know that a power series for a function must be the Taylor series (we did this is lecture #12, on March 1). Therefore the a_n's must be [f⁽ⁿ⁾(z₀)]/n!, and we have proved what is usually called
The Cauchy Integral Formula for Derivatives
f⁽ⁿ⁾(z₀)=[n!/(2Pi i)]_|w-z0|=r[f(w)/(w-z)ⁿ⁺¹]dw.

A hard theorem
Here is a totally silly result. I know of no application of this result.
IF you have a function f(z) defined on all of the complex numbers (all of C) with the following ludicrous property:
For any point z₀ in the plane, f(z) is equal to the sum of a power series centered at z₀, and this power series has radius of convergence equal to 1.
THEN the radius of convergence of f(z)'s Taylor series centered at any point is infinite, and f(z) is equal to the sum of that power series.

If you think about this result, it is very puzzling. Somehow some "thing" is pushing the radius of convergence bigger and bigger and bigger. I don't know any simple direct proof of this result. But let's use the machinery that we have:
Since f(z) is a sum of a power series near every point, f(z) is analytic in all of C. Since f(z) is analytic in all of C, then I can take the r in the picture/proof above as big as I like and the darn series will have infinite radius of convergence. So I'm done!
Below is a suitably silly picture illustrating this "result" (?).

Counterexample (?)
(At least for R!) I claimed that 1/(1+x²) was a function which was equal to a power series valid in some interval around every point. I will talk more about this in the next class.

Zero integrals imply .... (Morera's Theorem)
Suppose I have a continuous function f(z) in a domain, and I know that for any rectangle whose inside is contained in the domain, the integral of f(z) around the boundary of the rectangle is 0. Well, if you look back, you can see that this condition implies that f(z) has an analytic primitive or antiderivative: there is an analytic function F(z) whose derivative is f(z). But ... but ... the derivative of an analytic function is analytic. We have just proved
Morera's Theorem: If f(z) is a continuous function and and if the integral of f(z) around the boundary of any rectangle whose inside is contained in the domain is 0, then f(z) is analytic.
This is nice because integrals are usually much easier to handle than derivatives. Integrals are averages, and derivatives are quotients of tiny numbers, so derivatives are more unstable than integrals almost by their nature. Weird, weird ...

My empire stretches forever!
There are almost surely no more major proof techniques to be shown in this course. The bag of tricks is empty. Of course, we will use the techniques in many different specific applications. But ... we have obtained a wonderful result:

The following characterizations of a function are logically equivalent:
The initial definition The limit limh-->0[f(z+h)-f(z]/h exists for all z in f(z)'s domain, and this limit (called f´(z)) is a continuous function of z. Goursat proved that continuity is not necessary.
The Cauchy-Riemann equations f(z)=u(x,y)+iv(x,y), where the functions u and v are real-valued and continuously differentiable, and these equations are valid in the domain:    ux=vy    uy=-vx
Morera's Theorem f(z) is continuous, and the integral of f(z) around the boundary of any rectangle whose inside is contained in the domain is 0.
Power series f(z) is defined in some domain, and, for any z0 in the domain, f(z) is equal to the sum of some power series centered at z0 with a positive radius of convergence.
Harmonic functions & friends Again, f(z)=u(x,y)+iv(x,y) with u and v real-valued. Then u and v are harmonic functions, and v is a conjugate harmonic function for u.

There are other equivalences, some of which I will discuss later. To me the nicest equivalence yet undisclosed has to do with conformality which I have mentioned earlier.

Liouville's Theorem
An entire function f(z) (analytic in all of C) which is bounded (so there is a [positive] real constant M with |f(z)|<=M for all z) must be constant.
Yet another counterexample (?)
sin(x)? Well, not really. sin(z) if z=iy is actually isinh(y). So sin(z) grows exponentially along the imaginary axis.
I proved this by looking at f´(z) with the CIF for derivatives and showing (via ML) that this must be 0. The proof is in the text.

On Thursday
I will lecture from 6:10 to 7:10, take a 10 minute break, and then lecture for another hour. I would like to do this for the next few Thursdays and catch up with the other section.

HOMEWORK
Please hand in these problems on Thursday, March 24:
2.3: 1, 2, 4, 9, 10, 14.

Maintained by greenfie@math.rutgers.edu and last modified 1/18/2005.