Let $\Omega \subset \mathbb{C}$ be an open set that contains the unit ball $D$ and let $f \in \mathcal{O}(\Omega)$ a non constant map s.t. $|f(z)| = 1$ for all $z \in \partial D$. Show that $f$ has a zero in $D$.
Here's my reasoning:
$f \in \mathcal{O}(D) \Rightarrow f $ satisfies the maximum principle, i.e. the maximum of $|f(z)|$ is on $\partial D$.
Suppose $f$ doesn't have zeros within $D$. I would like to reach a contradiction with the fact that $f$ is not constant.
We would have: $\frac{1}{f} \in \mathcal{O}(D)$, so $1 \over f$ satisfies the maximum principle as well, thus we can conclude that $|f(z)| = 1$ for all $z \in D$.
But now I'm stuck, I don't know how to conclude that $f$ is indeed constant…any help is appreciated!
The maximum modulus principle implies that the maximum of $|f(z)|$ can only occur on the boundary. Another way to say it: for all $z\in D^\circ$, we have $|f(z)|<\max_{z\in\partial D}|f(z)|$. Since you have $|f(z)|=1$ for all $z\in D$, $f(z)$ must be constant.