Let $D$ be an open set in the complex plane and $f(z)$ be a non-constant holomorphic function on D. Then $|f(z)|$ has no local maximum on D.
I can follow the proof fine - usually if I don't understand a theorem intuitively beforehand, the proof will offer the insight necessary. Here, however, I can't see the reason for the Maximum Principle to hold - or perhaps I was just too shallow in my grasp of the proof. Does anybody have any shillings of wisdom that they would be willing to offer? Cheers.
On
holomorphic means, among other things, that the map is open. This is immediate when $f'(z_0) \neq 0,$ the Inverse Function Theorem says that an open neighborhood of $f(z_0)$ is covered surjectively. Even when $f'(z_0) = 0,$ the surjective part still holds, it is just that the map is locally $k$ to one, where $k$ is the first derivative such that $f^{(k)} (z_0) \neq 0.$ So, there it acts the way $z^3$ acts around the origin, for example.
Anyway, your $D$ is open; assume that the modulus takes its maximum at some point $z_0 \in D.$ Well then $f$ maps a neighborhood of $z_0$ onto a neighborhood of $f(z_0),$ including points with larger modulus than $f(z_0)$
Think about what the mean value property of analytic functions says: $f(z_0) = \dfrac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) d\theta$, where $f$ is analytic in the disk $B_r(z_0)$. This says that $f$ is equal to the average of the boundary points. How can $|f(z_0)|$ be larger than every single point of which it is the average? Thinking discretely, if $a= \dfrac{a_1 + \cdots + a_n}{n}$, can $a$ be larger than every point in this sum? No, this is not possible.