# UMD Analysis Qualifying Exam/Jan09 Complex

## Problem 2

edit## Solution 2

edit## Problem 4

edit
Suppose that are analytic on with on . Prove that for all implies |

## Solution 4

edit### Define new function h(z)

editDefine .

#### h is continuous on the closure of D

editSince on , then by the Maximum Modulus Principle, is not zero in .

Hence, since and are analytic on and on , then is analytic on which implies is continuous on

#### h is analytic on D

editThis follows from above

### Case 1: h(z) non-constant on D

editIf is not constant on , then by Maximum Modulus Principle, achieves its maximum value on the boundary of .

But since on (by the hypothesis), then

on .

In particular , or equivalently

### Case 2: h(z) constant on D

editSuppose that is constant. Then

where

Then from hypothesis we have for all ,

which implies

Hence, by maximum modulus principle, for all

i.e.

Since , we also have