# Show that if $f'(z)=0$ in a connected open set in $\mathbb{C}$, then $f(z)=a \in \mathbb{C}$ [duplicate]

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?

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

• Use the Mean value theorem. – Forever Mozart Feb 15 '16 at 5:03
• 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... ;-) – bartgol Feb 15 '16 at 5:22
• Is $a$ real number? How about $f(z)=i$ in $\mathbb{C}$? – choco_addicted Feb 15 '16 at 5:27
• 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. – user1050421 Feb 15 '16 at 5:36
• But the proposition doesn't say that $a$ is real. – choco_addicted Feb 15 '16 at 5:41

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 $z\in R$.
• Hmm... $f(z)$, I suggested, has derivative $0$, and $R$ is an open set, but not a constant. Is it wrong? – choco_addicted Feb 15 '16 at 5:54