If $|f(a)| \leq |f(z)| \forall z \in G$ then either $f(a) =0 $ or $f$ is constant.

Question

Let $G$ be a region and suppose that $f : G \rightarrow \mathbb{C}$ is analytic and $a \in G$ such that $|f(a)| \leq |f(z)| \forall z \in G$

Then either $f(a) =0 $ or $f$ is constant.

I concluded that $f$ is constant by applying Liouville's theorem by taking $|f(a)| $ to be some $M$ and showng that $0 \leq |f(z)| \leq \frac{1}{M}$ , showing $f$ is bounded and hence $f$ is constant throught out $\mathbb{C}$ , but i am not able to prove the part that $f(a) = 0$, any idea guys !

Thanks!

Answer 1

I'm not sure why positrón0802 deleted the answer.

If $f$ is analytic and non constant on $G$, it is an open map.

Suppose $f$ is non constant.

Since $a \in G$, if $f(a) \neq 0$, then $f(G)$ is an open set containing $f(a)$, hence it contains $t f(a)$ for some $t \in (0,1)$ and since $|tf(a)| < |f(a)|$, we have a contradiction.

Hence either $f(a) = 0$ or $f$ is constant.

Answer 2

You cannot apply Liouville's theorem because your function is not entire (holomorphic in all $\mathbb{C}$). This excercise can be solved by applying maximum principle. Consider $g(z)=\frac{1}{f(z)}, z\in G$ .Then it is true that if $f(a)\ne 0 $,then $$|g(z)|\le|g(a)|$$ for every $z\in G.$ That is $g(z)$ is constant inside $G$. Hence, either $f(a)=0$ and you cannot have any other conclusions, or your function is constant .