# UMD Analysis Qualifying Exam/Aug12 Complex

## Problem 4

 Suppose ${\displaystyle h(z)}$  is holomorphic on a region containing the disk ${\displaystyle \{|z|\leq R\}}$  and that ${\displaystyle |h(z)|  if ${\displaystyle |z|=R}$ . How many solutions does the equation ${\displaystyle h(z)=z}$  have in the disc ${\displaystyle \{|z|\leq R\}}$ ? Justify your answer.

### Solution 4

We know ${\displaystyle |h(z)|=|h(z)-z+z|  on ${\displaystyle |z|=R}$ . Similarly, since ${\displaystyle |h(z)-z|\geq 0}$  then ${\displaystyle |h(z)-z|+|z|\geq R}$  on ${\displaystyle |z|=R}$ . This gives ${\displaystyle |h(z)-z+z|  on ${\displaystyle z=R}$ .

So by Rouché's theorem, since both functions are holomorphic (i.e. have no poles), then ${\displaystyle h(z)-z}$  has the same number of zeros as ${\displaystyle z}$  on the domain ${\displaystyle |z| . Since ${\displaystyle z}$  has only one zero (namely 0), then there is only one solution to ${\displaystyle h(z)=z}$  inside the open disc ${\displaystyle |z| .

Observe that ${\displaystyle h(z)\neq z}$  for any ${\displaystyle |z|=R}$ , since that would imply ${\displaystyle |h(z)|=R}$  for some ${\displaystyle z}$  on the boundary, contradicting the hypothesis.

Thus, there is only one solution to ${\displaystyle h(z)=z}$  inside the open disc ${\displaystyle |z|\leq R}$ .