Prove that if $|f(z)| \geq |f(z_{0})|$ then $f(z_{0})=0$ Let $U \subseteq \mathbb{C}$ be an open connected subset and $f: U \rightarrow \mathbb{C}$ an holomorphic function.
Let $z_{0} \in U$ and $r > 0$ with $B_{r}(z_{0}) \subseteq U$.
If $\forall z \in B_{r}(z_{0})$ $|f(z)| \geq |f(z_{0})|$ then $f(z_{0})=0$.
As $f$ is an holomorphic map $f$ has the mean value property and then
$|f(z_{0})|\leq \max_{\partial B_{r}(z_{0})}|f|$
but I don't how to derive that is zero.
Thanks in advance.

If $f(z_0)\not=0$ consider an open connected nbd $V$ of $z_0$ in which $f$ is never vanishes and then apply maximum modules principal to $g:=\frac{1}{f|_V}$. That's $|g|\leq |g(z_0)|$, so $g$ is constant on $V$. So identity theorem implies $f\equiv f(z_0)$ on $U$. 
