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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

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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.

Seirios
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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?

Jonas Meyer
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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!)

Eric Auld
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    If the assumption is that the derivative is $0$, there is no problem using that the derivative is continuous. – Jonas Meyer Jul 09 '13 at 06:51
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    To consider points between $x$ and $y$, would you want to use convexity? Regardless you describe a way to show that $f$ is constant in a neighborhood of each point which could be used in combination with connectedness to show it is constant. – Jonas Meyer Jul 09 '13 at 06:55
  • @JonasMeyer My idea (similar to what you say) is to take any two points and use a polygonal path between them. – Eric Auld Jul 09 '13 at 06:56
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    I see. Similarly, one could use that $u_x=u_y=0$ implies that $u$ is constant on each horizontal or vertical segment, and between any two points use a polygonal path where each edge is horizontal or vertical. – Jonas Meyer Jul 09 '13 at 06:59