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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?

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    $\begingroup$ I don't understand why they don't say the maximum modulus and open mapping are easy consequences of the holomorphic $\implies$ analytic theorem (the latter means $f(z) = f(z_0) + C(z-z_0)^n+O((z-z_0)^{n+1})$) $\endgroup$ – reuns Dec 8 '18 at 20:56
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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.

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