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Suppose $\Omega$ is a bounded domain in the complex plane whose boundary consists of $m+1$ disjoint analytic simple closed curves. Let $f$ be a non-constant holomorphic function on a neighborhood of $\overline\Omega$ such that $|(f(z)|=1$ on the boundary. I wish to prove that $f$ must have $m+1$ zeroes in $\Omega$.

I am trying to follow the solution given here, but I am running into problems. In particular, if $C$ is a curve on the boundary, I do not see why $f$ changes argument by at least $2\pi$ on $C$. The linked solution says this follows from $f(C)$ being the unit circle, but I do not see why this is. Why couldn't $f$ trace out the unit circle clockwise, and then return counterclockwise? In that case the image is the unit circle, but it doesn't wind around. In fact, I don't even see why the image must all of the unit circle.

I feel that the justification for this must use the other properties of the problem. It doesn't work just on arbitrary individual curves. For example, consider the automorphism of the unit disk moving $1/2$ to $0$ and sending the boundary of the unit disk to itself. Call this $f(z)$. Consider $f(z)/z$ on the same region and remove a small circle around $0$. This is an analytic function on the annulus, it is of constant modulus $1$ on the boundary, but when we integrate along the unit circle we get a total change in argument of 0. Note that this even admits an analytic extension to a neighborhood of this closed region.

Any help would be great. I would appreciate as much detail as possible. I also have a bounty on the same question posed here, so if you can answer this, please also answer that and claim the bounty. I feel this question better explains my thoughts on the problem than a comment could, which is why it is posted as a separate question.

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Can't you proceed like this : Fix some point $z_0$ in the boundary of $\Omega$, then consider the holomorphic function $g(z)=\log(f(z))=\log|f(z)|+i \operatorname{arg}(f(z))$ near $z_0$. Note that $\log|f(z)|<0$ in $\Omega$ and vanishes on the boundary. Applying the Cauchy-Riemann equations to $g$, we obtain that $\operatorname{arg}(f(z))$ is strictly increasing near $z_0$, as $z$ moves on the curve. –  Malik Younsi Dec 21 '12 at 21:03
    
Do you have an example of such an $f$ for the simplest case of an annulus $r <|z|<R$? –  WimC Dec 21 '12 at 21:42
    
@WimC : See my answer to this question : math.stackexchange.com/questions/95570/… I do now know any "explicit" example though. –  Malik Younsi Dec 21 '12 at 22:43
    
@Potato : Your example does not satisfy $|f(z)|\equiv 1$ on the boundary of $\Omega$, which is essential to make sure that $f$ maps $\Omega$ in the unit disk (which follows from the maximum principle..) –  Malik Younsi Dec 21 '12 at 22:45
    
@MalikYounsi Thanks for the pointer. I think that explicit examples can be constructed from elliptic functions or theta functions of degree $4$ for an orthogonal lattice. –  WimC Dec 22 '12 at 7:09

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For a proof of this fact, see the end of the proof of theorem 4.5.9. in "Geometric function theory: explorations in complex analysis", by Krantz.

The proof is the same than the one I gave in the comments.

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