$G$ open connect subset of $\mathbb{C}$ and $f: G \to \Bbb C$ analytic. 
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*$G$ open connect subset of $\mathbb{C}$ and $f: G \to \Bbb C$ analytic. 

*Suppose $\exists a \in G$ such that $|f(a)| \leq |f(z)|$ for all $z \in G$. 

Prove that either $f(a) = 0$ or $f$ is nonzero constant in $G$.

Suppose $f(a) \not = 0$ and $f$ is not a non zero constant function on $G$. 
I just need a hint then I think I should be able to finish the problem.
 A: Hint:  use the maximum modulus principle or open mapping theorem.
Solution: Recall that $f(z)$ is an open mapping unless it is constant (by the open mapping theorem). This implies that a non-constant mapping maps $G$ to an open subset of $\Bbb C$, and $|\cdot |$ maps this to an open subset of $\Bbb R_{\ge 0}$ (this is how one proves the maximum modulus principle). Since $G$ is connected, $f(G)$ is also connected, hence $f(G)=(c,d)$ or $[0,b)$ for some $0<c<d\in\Bbb R$ or $0<b\in\Bbb R$ since these are the only open subsets of $\Bbb R_{\ge 0}$. In the case $f(G)=(c,d)$ there is no minimum, so since we assumed $f$ has one and is non-constant, it must be that $f(G)=[0,b)$, i.e. the minimum is $0$.
If you already know the maximum modulus principle then note that if $|f|>0$ then ${1\over f}$ achieves a maximum, hence $f$ is constant.
A: The things you say for both directions aren't really sensible.
First note that there is obviously no problem with $f$ being constant.  So that deals with that case.
Now suppose that $f$ is not a constant and that $f(a)\neq 0$.  Derive a contradiction.
