Zeros of a holomorphic function Suppose $\Omega$ 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 $\Omega$ such that

$$|f(z)|=1$$ for all $z$ in the boundary of $\Omega$.

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

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

Thank you
This is true and it is a consequence of the Argument Principle; the Maximum Modulus Principle is not precise enough to prove it. Here is the proof for $m=0$ and it generalizes: We know that $$n(w)=\frac{1}{2\pi i}\int_{\partial \Omega}\frac{f&#39;(z)}{f(z)-w}\, dz$$ gives the number of times $f$ equals $w$ in $\Omega$ and is continuous on $D=\{|w|&lt;1\}$. Since $f$ is not constant, $n(w)\ge 1$ for some $w\in D$. By connectedness and continuity, we also have $n(0)\ge 1$. Therefore $f$ has at least one zero in $D$.

In the general case, let $\partial\Omega=\gamma_1\cup\cdots\cup \gamma_{m+1}$ with orientation $\gamma_1-\cdots- \gamma_{m+1}$ (so $\gamma_1$ is also the boundary of the unbounded compoent of ${\mathbb C}\backslash\Omega$). Using the Open Mapping Theorem, one can deduce that each $f(\gamma_k)$ traverses the unit circle at least once, with $f(\gamma_1)$ in the positive direction and $f(\gamma_k)$ in the negative direction for $k\ge 2$. So each point on the unit circle is assumed by $f$ at least $m+1$ times on $\partial\Omega$. Use this fact and a point $w$ close to the unit circle to see that there are at least $m+1$ points in $\Omega$ that has image $w$. So $n(w)\ge m+1$. Now use the same argument as the $m=0$ case to conclude that $n(0)\ge m+1$. (Or one can just use an analytic continuation of $f$ and do away with $w$ close to the unit circle.)
Since each of the $m+1$ simple closed curves are disjoint, it is clear that $\Omega$ has $m+1$ connected components. Let $U$ be any one of these.

Then $f(z)$ restricted to $U$ is a holomorphic function. Furthermore, it is non-constant and holomorphic on some neighborhood of the closure of $U$, because this closure is contained in $\Omega$. Furthermore, the boundary of $U$ is contained in the boundary of $\Omega$, so $|f(z)| = 1$ on the boundary of $U$.

Like you outlined in your explanation of the problem, if we assume that $f$ is non-zero in $U$ then both $f(z)$ and $1/f(z)$ are holomorphic in $U$, and thus must achieve their maximums on the boundary of $U$. But its given that $|f(z)| = 1$ on the boundary of $U$, and so we get

$|f(z)| \leq 1$ and $|1/f(z)| \leq 1$ for $z \in U$. Thus we must have that $|f(z)| \equiv 1$ on $U$, which is a contradiction as we assumed that $f(z)$ is non-constant. Thus $f$ has at least one zero in $U$. We assumed nothing about which connected component we used, so this is true in each component of $\Omega$, and therefore $f(z)$ has at least $m+1$ zeroes.