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Suppose Ω is a bounded domain in the plane whose boundary consist of m+1 disjoint analytic simple closed curves.

Let f be holomorphic and nonconstant on a neighborhood of the closure of Ω such that

|f(z)|=1 for all z in the boundary of Ω.

If m=0, then the maximum principle applied to f and 1/f implies that f has at least one zero in Ω.

What about the general case? I believe that f must have at least m+1 zeros in Ω, but I'm not able to prove it...

Thank you

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1 Answer 1

Let $C$ be a curve among the m+1 curves defining the boundary. Then since $f$ is non-constant and analytic, $f(C)$ equals the unit circle. Hence $\frac1{2\pi}\Delta_C \arg f(z) \ge 1$. Doing this for all the curves we have $\frac1{2\pi}\Delta_{\partial \Omega} \arg f(z) \ge m+1$ and so $f$ has $m+1$ roots in $\Omega$ by the argument principle.

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To translate this into topological language: $f$ is a map from $\Omega$ to the unit disk that takes boundary to boundary. (The image of $\Omega$ is in the disk by the maximum principle). The number of zeros (counted with multiplicity) is the degree of the map. The degree of the map equals the degree of the map on the boundary, which must be at least $m+1$ (since analyticity forces the map to be monotone increasing on each boundary component) –  Bill Thurston Jan 3 '11 at 16:20
Thank you, but could you elaborate a little on why $f(C)$ equals the unit circle? –  Analyst2 Jan 4 '11 at 16:02

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