TITLE: Holomorphic function with zero derivative is constant on an open connected set
QUESTION [5 upvotes]: I was wondering about this fact, as I do not know how to prove it correctly. I tried with Cauchy-Riemann, but since they are PDE's I found it hard to show that this is the only thing that can cause this zero derivative

REPLY [13 votes]: By integration: Let $\Omega \subset \mathbb{C}$ be a connected open set, $f : \Omega \to \mathbb{C}$ be a holomorphic function and $x_0 \in \Omega$. For every $x \in \Omega$, there exists a path $\gamma : [0,1] \to \Omega$ such that $\gamma(0)=x_0$ and $\gamma(1)=x$.
$$0= \int_{\gamma} f'(z)dz= \int_0^1f'(\gamma(t)) \gamma'(t)dt= \int_0^1(f \circ \gamma)'(t)dt= f(x)-f(x_0)$$
Therefore, $f(x)=f(x_0)$ for every $x \in \Omega$, ie. $f$ is constant.