prove an analytic function has at least n zeros 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...

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.
