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I understand that the Maximum Modulus Principle works, but I'm a little baffled as to why. To be more precise, the picture I have in my head is something like this: for a compact set $K \subset \mathbb{C}$, since $|f|$ (for $f$ holomorphic) can only attain its maximum on the boundary $\partial K$, if you consider a disk centered on the origin, no matter what value $|f|$ obtains on the boundary of this disc, you can just increase the radius of the disc a little bit and find a higher value of $|f|$; in other words, the absolute value of function just keeps growing without bound. What exactly is driving/forcing this growth?

I'm looking for some kind of explanation, geometric or otherwise, that could aid my intuition. In particular, why does $\mathbb{C}$ behave so differently from $\mathbb{R}$ here?

EDIT. In response to the "duplicate" tag: I am looking for something a little deeper than the answers there. As mentioned above, an explanation of the difference in behaviours of $\mathbb{C}$ and $\mathbb{R}$ in this regard, perhaps with a tie-in to the Cauchy-Riemann equations...it seems like there's something that causes functions to behave fundamentally differently over $\mathbb{C}$, and I would like to understand why, with the Maxiumum Modulus Principle as a concrete example.

Anthony
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In either case $$f(z+h)=f(z)+hf'(z)+\dots,$$where the dots indicate higher order terms that are smaller than the last term if $h$ is small. Now if $f'(z)\ne0$ this shows that $|f|$ cannot have a maximum at $z$, because we can choose $h$ to point in the right direction so $|f(z)+hf'(z)|>|f(z)|$.

But what if $f'(z)=0$? Then $$f(z+h)=f(z)+\frac12h^2f''(z)+\dots.$$In the real case $h^2\ge0$, so if $f''(z)$ has the opposite sign to $f(z)$ then adding the $h^2f''(z)$ makes $|f|$ smaller. But in the complex case $h^2$ can point in any direction; so if we choose the direction for $h^2f''(z)$ to be the same as the direction of $f(z)$ then adding the $h^2f''(z)$ again makes $|f|$ larger.

Hmm, slightly more formally: If $f'(z)\ne0$ then $z$ cannot be a maximum for $|f|$, for very much the same reason in the real and complex case. Suppose that $f'(z)=0$, $f''(z)\ne0$. Now in the real case, if $f(z)>0$ and $f''(z)<0$ then $$|f(z)+\frac12 h^2f''(z)| =|f(z)|-\frac12h^2|f''(z)|<|f(z)|$$ for all small $h\ne0$, regardless of whether $h$ is positive or negative, the only two directions available. But that can't happen in the complex case: If $f''(z)\ne0$ then there exists $\alpha$ so that if $h=re^{i\alpha}$ then $$|f(z)+\frac12h^2f''(z)|=|f(z)|+r^2|f''(z)|>|f(z)|$$ for all small $r>0$.

David C. Ullrich
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