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Show that if $f'(z)=0$ from the function $f : D \to \mathbb{C}$ which $D \subset \mathbb{C}$ is a domain (i.e. open connected set), then $f(z)=a \in \mathbb{C}$ ($a$ is a constant in $\mathbb{C}$). The conclusion is it the same if $f'(z)=0$ on any open set?

I think I can use the Cauchy-Riemann theorem, but it's unclear.

Is anyone could explain to me how to solve this problem?

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marked as duplicate by egreg, Jack's wasted life, N. F. Taussig, hardmath, Em. Feb 15 '16 at 12:49

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    $\begingroup$ Use the Mean value theorem. $\endgroup$ – Forever Mozart Feb 15 '16 at 5:03
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    $\begingroup$ I believe this is not a function of complex variables (otherwise the conclusion would be false). If so, then Cauchy-Riemann has no meaning here. Also, I would reserve the letter $z$ for complex variables. Not that it is necessary, but it's like using $n$ for a real variable... ;-) $\endgroup$ – bartgol Feb 15 '16 at 5:22
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    $\begingroup$ Is $a$ real number? How about $f(z)=i$ in $\mathbb{C}$? $\endgroup$ – choco_addicted Feb 15 '16 at 5:27
  • $\begingroup$ A proof exist from this website : home.ku.edu.tr/~bozbagci/401Spring14M1-solutions.pdf. However, I want a proof in using the Cauchy-Riemann theorem. $\endgroup$ – user1050421 Feb 15 '16 at 5:36
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    $\begingroup$ But the proposition doesn't say that $a$ is real. $\endgroup$ – choco_addicted Feb 15 '16 at 5:41
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If $f'(z)=0$, then, every directional derivative of $f(z)$ is $0$. Thus, $$ \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=0, $$ where $u(x,y)$ and $v(x,y)$ are real and satisfy $f(x,y)=u(x,y)+iv(x,y)$. Comparing coefficients, we get $\frac{\partial u}{\partial x}=0$ and $\frac{\partial v}{\partial x}=0$. Then by Cauchy-Riemann equation, $$ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}=0,\qquad \frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=0. $$ Then follow Problem 1(a) of the website you showed. Or... If $f(z)$ analytic in the domain $D$ and $f'(z)=0$ so $f(z)=$constant? will also help you.


Let $R$ be a union of $B(0,1)$ and $B(2,1)$, where $B(a,r)$ is a ball whose center is $a$ and radius is $r$. That is, $$ B(a,r)=\{z\in \mathbb{C}:|z-a| < r\}. $$ Since $B(0,1)$ and $B(2,1)$ are open sets, their union is also an open set. Define $f:R\to\mathbb{C}$ such that $$ f(z)=\begin{cases} 1,& z\in B(0,1)\\ 2,& z\in B(2,1) \end{cases}, $$ then $f'(z)=0$ for all $z\in R$.

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  • $\begingroup$ Two balls in example are not connected. $\endgroup$ – runaround Feb 15 '16 at 5:47
  • $\begingroup$ @runaround I intended to show that the latter proposition is not true. $\endgroup$ – choco_addicted Feb 15 '16 at 5:51
  • $\begingroup$ Hmm... $f(z)$, I suggested, has derivative $0$, and $R$ is an open set, but not a constant. Is it wrong? $\endgroup$ – choco_addicted Feb 15 '16 at 5:54
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    $\begingroup$ I'm editing my post, but I feel you are too offensive to threat me. Can't you wait a few minutes? $\endgroup$ – choco_addicted Feb 15 '16 at 6:08

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