Solutions to Dr. Z.'s Intro to Complex Variable Attendance Quiz for Lecture 21



1.    State and Prove the Maximum Modulus Principle.

(Hint: you can use the intuitively obvious fact that if f(z) is a non-constant analytic function on
a domain D, then the range of f(D) is an open set).


Sol. to 1: 
STATEMENT: The  Maximum Modulus Principle says that if f(z) is a  nonconstant analytic function
on a domain D, then |f(z)| can have no local maximum on D.
In other words the maxima of |f| on the CLOSURE of D must be on the BOUNDARY of D.

PROOF:
Suppose,  on the contrary, that there is a local maximum z=z0  INSIDE D.
Since D is OPEN (that part of the definition of DOMAN), it means that there is SOME positive number r (possibly very small) such that
the disc |z-z0| <r is contained in D and |f(z_0)| >= |f(z)| for the all points z inside that disc.

But this means that f(z0) lies on the BOUNDARY of the IMAGE of  disc  |z-z0| <r under the
function f(z), namely the open set

W= { f(z)  :  |z-z_0|<r}


But W is an open set containing f(z0), contradiction!


2. Let f(z)=2*z^3/(z+2)^4 .  Find  the maximum value  of |f(z)| as z varies over the disc |z|<=1 .



 Sol. to 2: By the  MAXIMUM PRINCIPLE, the maximum VALUE of |f(z)| must lie on the
BOUNDARY of the disc  |z|<=1. In other words on the  CIRCLE |z|=1.

On that disc

|f(z)|= 2|z|^3/|z+2|^4= 2/|z+2|^4


Recall that |z+2|=|z-(-2)| is the  DISTANCE of z from the point -2 (alias (-2,0)). The   SHORTEST
distance is acheived at the point z=-1, (alias (-1,0)), and that distance is 1 
(draw a picture!). Hence the minimum of |z+2| as z goes over |z|=1 is 1, hence the  MAXIMUM of the
reciprocal 1/|z+2|^4 is  1 and the MAXIMUM of   1/|z+2|^4 is 1. Plugging it in, we have:

 Ans. to 2.: The maximum value  of |f(z)| as z varies over the disc |z|<=1 is 2.