# $|f(z)|=c\in\Bbb R\forall z\in\Omega$ for $\Omega\subset\Bbb C$ open and connected $\implies f$ is constant. Why must $\Omega$ be open, connected?

Let $$f$$ be holomorphic on $$\Omega\subset\Bbb C$$, and $$\Omega$$ is open and connected. If $$|f(z)|$$ is constant it is known that $$f$$ is constant on $$\Omega$$

The proof of this fact says that if $$f$$ is not constant then $$f(\Omega)$$ is open as the image of an open by a holomorphic function. But a circle of radius $$c\in\Bbb R_+$$ is not open and $$f(\Omega)$$ wouldn't be open either as a subset of that circle, which contradicts the first sentence.

Question 1

Where is the connectedness of $$\Omega$$ used in the above proof?

Question 2

Why should $$\Omega$$ have any of the two constraints? Is the following proof invalid?

writing $$z=x+iy$$ and $$f(z)=u(x,y)+iv(x,y)$$ since $$f$$ is holomorphic the Cauchy-Riemann equations hold: $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}~~\text{ and }~~\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ $$\forall (x+iy)\in\Omega\ ~|f(x+iy)|=\sqrt{u^2(x,y)+v^2(x,y)}=c\in\Bbb R_+$$

$$\implies\frac{\partial }{\partial x}[\sqrt{u^2(x,y)+v^2(x,y)}]=\frac{\partial }{\partial y}[\sqrt{u^2(x,y)+v^2(x,y)}]=0$$

$$\iff (u_x+v_x)(u^2+v^2)^{-{1\over2}}=(u_y+v_y)(u^2+v^2)^{-{1\over2}}=0$$

If $$u^2+v^2=|f|^2=0\forall z\in\Omega$$ then $$f=0$$ and we're done

Let's suppose $$u^2+v^2\ne0$$ then we can divide by $$(u^2+v^2)^{-{1\over2}}$$ it on both sides to get: $$u_x+v_x=u_y+v_y=0$$ by the C-R equations $$u_x-u_y=u_y+u_x\implies u_y=-v_x=0$$ substituting $$v_x=0$$ into $$u_x+v_x=0$$ $$u_x=0\implies v_y=0$$

All the partial derivatives of $$u$$ and $$v$$ are zero and since $$f'(z)=u_x(x,y)+iv_x(x,y)=0$$ we get that $$f$$ is constant.

• In your proof, you use the fact of connectedness of domain: If a multivariable function has its total derivative zero in a connected domain, then the function is constant on that domain – vidyarthi Dec 18 '18 at 17:09
• Let $f=1$ on $B(0,1)$ and $f=-1$ on $B(2,1)$ then $|f|$ is constant but $f$ is not. You need connectedness so the constant is the same. – copper.hat Dec 18 '18 at 17:12
• The conditions are sufficient, not necessary. $f$ may be constant without $\Omega$ being either open or connected. – copper.hat Dec 18 '18 at 17:13
• But the connectedness is necessary, no? – John Cataldo Dec 18 '18 at 17:14
• You were asking in Question 1 where connectedness is used, I am giving an example where $\Omega$ is not connected, but is open and $|f|$ is constant but $f$ is not. This should indicate where connectedness is used. – copper.hat Dec 18 '18 at 17:26