# Holomorphic functions with constant modulus

Problem Statement: Suppose that $f$ is holomorphic on a domain $\Omega \subseteq \mathbb{C}$ with $|f|$ constant. Prove that $f$ is a constant map.
Ideas: If $|f| = 0$ then $f = 0$. Suppose $|f| = b \in \mathbb{C}$, $b \ne 0$. So $f(z) = b e^{i \theta(z)}$ where $\theta(z) = arg_0(f(z))$ with $arg_0(f(z))$ is a holomorphic branch of $arg(f(z))$; if such a branch did not exist then $f$ would not be holomorphic on $\Omega$. I tried applying the definition of the complex derivative directly to show that $f' = 0$ to no avail. If it were shown that for any $z_1 \ne z_2$ in $\Omega$ that $\theta(z_1) = \theta(z_2)$ it would be done, but I don't see how to make that work either.
If you want a one-liner use the open mapping theorem, otherwise use the C-R equations on the identity $u^2+v^2=|f|^2=c$ where $f(z)=u(z)+iv(z)$.