Holomorphic function with zero derivative is constant on an open connected set 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
 A: If $f(x+iy)=u(x,y)+iv(x,y)$, then $f'(x+iy)=\lim\limits_{h\to0,h\in\mathbb R}\dfrac{f(x+iy+h)-f(x+iy)}{h}=u_x(x,y)+iv_x(x,y)$, and $f'(x+iy)=\lim\limits_{h\to0,h\in\mathbb R}\dfrac{f(x+iy+ih)-f(x+iy)}{ih}=\dfrac{1}{i}(u_y(x,y)+iv_y(x,y))$.  
Hence you can conclude that the partial derivatives of the real and imaginary parts of $f$ vanish everywhere.
Given a function $u$ from a connected open subset of $\mathbb R^2$ to $\mathbb R$ with $u_x=u_y\equiv 0$, can you prove that $u$ is constant?
A: As Jonas alludes to, there is a theorem from multivariable calculus that if a real-valued function has continuous partial derivatives on an open set $U$, then for every $x,y\in U$ there exists a point $\eta$ strictly between $x$ and $y$, and lying on the line between them, such that $g(x) - g(y) = \langle \nabla g(\eta),x-y\rangle$. You can use this, treating $g$ as first the real part of your function then the complex part. So this will solve the problem, once we note that an open connected set in $\mathbb{C}$ is polygonally connected.
Note: For this you would need to use the fact that if a function is complex-differentiable, then its derivative is continuous (in fact, it is itself differentiable), but perhaps you have learned this. (edit: Jonas has reminded me that the derivative is zero, so of course it is continuous!)
A: 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.
