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.

  • $\begingroup$ 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 $\endgroup$ – vidyarthi Dec 18 '18 at 17:09
  • $\begingroup$ 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. $\endgroup$ – copper.hat Dec 18 '18 at 17:12
  • $\begingroup$ The conditions are sufficient, not necessary. $f$ may be constant without $\Omega$ being either open or connected. $\endgroup$ – copper.hat Dec 18 '18 at 17:13
  • $\begingroup$ But the connectedness is necessary, no? $\endgroup$ – John Cataldo Dec 18 '18 at 17:14
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    $\begingroup$ 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. $\endgroup$ – copper.hat Dec 18 '18 at 17:26

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