TITLE: If $f$ is analytic and $f(z)^2$ = $\bar f(z)$ then $f$ is constant
QUESTION [6 upvotes]: I'm currently stuck on the following problem. 
Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$\bar f(z)$ $\forall z\in V$ then f is constant on V. 
I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.

REPLY [2 votes]: $$\|f(z)\|^2 = \|f(z)^2\|=\|\overline{f}(z)\|=\|f(z)\| $$
so $\|f(z)\|$ is either $0$ or $1$, constantly, since $\|f(z)\|$ is continuous. In the former case $f(z)\equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $\xi^2=\overline{\xi}$, so our $f$ is constantly $0$, $1$, $\omega$ or $\omega^2$.