2
$\begingroup$

I`m confused about this problem: Let G be a bounded region in C whose boundary consists of n circles. Suppose that f is a non-constant function analytic on G: Show that if absolute value of f(z) = 1 for all z in the boundary of G then f has at least n zeros (counting multiplicities) in G.

What does it mean that boundary consists of n circles? How can I start solving the problem? Any help please...

  • 1
    See http://mathoverflow.net/questions/51029/zeros-of-a-holomorphic-function2012-10-24

1 Answers 1

2

I don't know what tools you have at your disposal, but this follows from some basic topology. The assumptions imply that $f$ is a proper map from the region $G$ to the unit disk $\mathbb{D}$. As such the map has a topological degree, and since the preimage of the boundary $|z|=1$ contains $n$ components, this degree has to be at least $n$, meaning that every $z\in\mathbb{D}$ has to have at least $n$ preimages, counted with multiplicity.

  • 0
    I definitely don`t know about that in topology, I meant to do that in complex analysis, but I`ll try to use what you said and see if that will help me.2012-10-23