TITLE: Zeros of a holomorphic function
QUESTION [6 upvotes]: 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

REPLY [9 votes]: 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.