Let $f$ be a non-constant holomorphic function in a domain $U$. Assume $z_0 ∈ U$ is a local minimum of $\lvert f\rvert$. Show that $f(z_0) = 0$. I have a question from my homework:

Let $f$ be a non-constant holomorphic function in a domain $U$. Assume $z_0 \in U$ is a local minimum of $|f|$. Show that $f(z_0) = 0$.

I know the maximum principle, but I can't figure out the minimum. I thought about Cauchy's integral for sub domain of $U$, but am unsure if that's any useful. 

Suppose $f(x_0) \neq 0$, then with $g(x) = {1 \over f(x)}$, the function $x \mapsto |g(x)|$ has a local maximum at $x_0$. Applying the maximum modulus theorem to a small ball around $x_0$ shows that $g$ (and hence $f$) is constant.
