Maximum Modulus Principle and Open Mapping Theorem

Question

The Maximum Modulus Principal states: let $f$ be a function holomorphic on some connected open subset $D$ of the complex plane $\mathbb{C}$ and taking complex values. If $z_{0}$ is a point in $D$ such that $|f(z_{0})|\geq |f(z)|$ for all $z$ in a neighborhood of $z_{0}$, then the function $f$ is constant on $D$.

The notes further state that: “alternatively, the maximum modulus principle can be viewed as a special case of the open mapping theorem, which states that a nonconstant holomorphic function maps open sets to open sets. If $|f|$ attains a local maximum at $z$, then the image of a sufficiently small open neighborhood of $z$ cannot be open. Therefore, $f$ is constant.

Can some explain why intuitively, if $|f|$ attains a local maximum at $z$, then the image of a sufficiently small open neighborhood of $z$ cannot be open?

Answer 1

If $z$ is a local maximum of $\vert f \vert$, then by definition for a sufficiently small open neighborhood of $U$ of $z$,

$y \in U \Longrightarrow \vert f(y) \vert \le \vert f(z) \vert; \tag 1$

this in turn implies

$\vert f \vert (U) \subset (\vert f(x) \vert - \epsilon, \vert f(z) \vert] \tag 2$

for some $\epsilon > 0$, which is not open since it does not contain an open set containing $\vert f(z) \vert \in \Bbb R$.

It is worth observing that only the continuity of $f$ is used here, not fact that $f$ is holomorphic.

Answer 2

$\newcommand{\abs}[1]{\left \lvert #1 \right \rvert}$

In general,

for any open set $U\subset\mathbb{C}$ , for any $z\in U$, there exists $w\in U$ s.t. $\abs{z}<\abs{w}$.

proof

We may assume $U=U(a,r)=\{z\in\mathbb{C}\mid \abs{z-a}<r\}\ (a\in\mathbb{C},r>0)$ and $z\neq 0$. Let $\epsilon:=\frac{r-\abs{z-a}}{2\abs{z}} (>0)$, $w:=(1+\epsilon)z$, then $\abs{w}>\abs{z}$ and $w\in U(a,r)$ because

$$ \abs{w-a}\leq \abs{z-a}+\epsilon\abs{z} =\frac{r+\abs{z-a}}{2} < r. $$