# prove an analytic function has at least n zeros

Im 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...

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The description of the boundary just means that $G$ is in $n$ pieces, and each one has a circle for its boundary. For instance the unit disk is a region with $1$ circle for its boundary; the union of the disks of radius $1/3$ centered at $i,2i,...,ni$ has for its boundary the $n$ circles of the same radius and the same centers. – Kevin Carlson Oct 23 '12 at 19:45
Region usually means that your set is connected, so $G$ consists of a disk with $n-1$ circular holes. An annulus would be an example with $n=2$. – Per Manne Oct 23 '12 at 22:32
– Kalim Oct 24 '12 at 17:22

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.
I definitely dont 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. – Danny Oct 23 '12 at 23:57