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I am currently reviewing theorems I already learned in complex analysis and looking at

Holomorphic function with zero derivative is constant on an open connected set

  1. I asked myself wether it is necessary that $f'(x)=0 \forall x \in G$ for $f:G\to \mathbb{C}$ and $G$ being a domain to get that $f$ is constant or wether a single point in the domain could be sufficient. If so, how to show that?

  2. How is this related to the maximum principle? When I think about real analysis, $f'(a)=0$ implies that there is an extremum in $a$ but it seems this is not the case in complex analysis, at least I have not found anything but the maximum principle regarding this issue.

Marsl
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  • Careful: $f(x) = x^3$ has derivative $0$ at $x=0$...no extremum. Similarly, you have things like saddle points for functions $\mathbb{R}^2 \rightarrow \mathbb{R}$. – Kaj Hansen Jun 10 '17 at 21:38
  • Right! Does this mean, if I have $f'(a)=0$ that $f$ need not be constant as $a$ could be just a saddle point? – Marsl Jun 10 '17 at 21:42
  • G must be connected, and a single point is not sufficient. – alexp9 Jun 10 '17 at 22:06

1 Answers1

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A single point where the derivative vanishes is not sufficient: consider $f(z)=z^2$ (defined on $\mathbb{C}$), whose derivative vanishes at $0$.

We can assume that the domain is connected; the main theorem is

Theorem. Suppose $f$ is holomorphic in the connected domain $G$ and that the set of zeros of $f$ has a limit point in $G$. Then $f$ is the zero function.

It follows that if $f$ is holomorphic on a connected domain $G$ and $f'(z)=0$, for all $z\in G$, then $f'(z)$ is the zero function (because it's holomorphic as well). Therefore $f$ is constant.

The maximum principle states that for a nonconstant holomorphic function defined on a domain $G$, the function $|f|$ has no local maximum inside $G$. This doesn't seem related to the vanishing of the derivative at some point.

egreg
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