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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...

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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.2012-10-23
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    *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 $n=2$.2012-10-23
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    See http://mathoverflow.net/questions/51029/zeros-of-a-holomorphic-function2012-10-24

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