Claim. Let $\Omega$ be an open connected subset of $\mathbb C$ and $\,f:\Omega\rightarrow\mathbb{C}$ is holomorphic. If $\lvert\, f\rvert$ is a constant function, then $f$ is also a constant function.
I tried to do this like this: Suppose $f=u+iv$. Then from the fact that $\lvert f\rvert=\sqrt{u^2+v^2}$ is constant we get
$$\frac{\partial\lvert f\rvert}{\partial z}=\frac{1}{2}\left(\frac{\partial \lvert f\rvert}{\partial x}+\frac{1}{i}\frac{\partial \lvert f\rvert}{\partial y}\right)=0$$
$$or,~~~2u\frac{\partial u}{\partial x}+2v\frac{\partial v}{\partial x}=0~~~and~~~2u\frac{\partial u}{\partial y}+2v\frac{\partial v}{\partial y}=0$$
But how do I proceed from here?
Also I need to know, What would happen if $\Omega$ were not connected?
Best Answer
If $f=u+iv$ and $$ u^2+v^2=c, $$ then differentiating by $x$, and using Cauchy-Riemann equations, we get $$ 0=uu_x+vv_x=uv_y-vu_y=(u_y,v_y)\cdot(-v,u) $$ and by $y$ $$ 0=uu_y+vv_y=(u_y,v_y)\cdot(u,v). $$ Thus the vector $(u_y,v_y)$ is parallel and perpendicular to $(u,v)\ne 0$, and thus $$ (u_y,v_y)=(0,0), $$ everywhere, and using Cauchy-Riemann again, we get that $(u_x,v_x)=(0,0)$, also everywhere.
Thus $u$, $v$ are constant.
Note. If $\Omega$ is not connected, then $f$ is constant in each connected component, but it might have a different value in each component.