# Holomorphic functions with constant modulus

Problem Statement: Suppose that $\displaystyle f$ is holomorphic on a domain $\displaystyle \Omega \subseteq \mathbb{C}$ with $\displaystyle |f|$ constant. Prove that $\displaystyle f$ is a constant map.
Ideas: If $\displaystyle |f| = 0$ then $\displaystyle f = 0$. Suppose $\displaystyle |f| = b \in \mathbb{C}$, $\displaystyle b \ne 0$. So $\displaystyle f(z) = b e^{i \theta(z)}$ where $\displaystyle \theta(z) = arg_0(f(z))$ with $\displaystyle arg_0(f(z))$ is a holomorphic branch of $\displaystyle arg(f(z))$; if such a branch did not exist then $\displaystyle f$ would not be holomorphic on $\displaystyle \Omega$. I tried applying the definition of the complex derivative directly to show that $\displaystyle f' = 0$ to no avail. If it were shown that for any $\displaystyle z_1 \ne z_2$ in $\displaystyle \Omega$ that $\displaystyle \theta(z_1) = \theta(z_2)$ 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 $\displaystyle u^2+v^2=|f|^2=c$ where $\displaystyle f(z)=u(z)+iv(z)$.