If $f$ is holomorphic and $\lvert\, f\rvert$ is constant, then $f$ is constant.

Question

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?

Answer 1

Continuing your approach, you have $u^2+v^2=r^2=const.>0$ and additionally to your derivatives of that equation you have the Cauchy-Riemann PDE, so that you get a regular homogeneous 4x4 system for the partial derivatives of $u$ and $v$, rendering them all zero.

$$\begin{bmatrix} u&v&0&0\\ 0&0&u&v\\ 1&0&0&-1\\ 0&1&1&0 \end{bmatrix} \cdot \begin{bmatrix} ∂_xu\\∂_xv\\∂_yu\\∂_yv \end{bmatrix}=0 $$

Answer 2

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.

Answer 3

If Ω is not connected, you cannot use that, if ux=uy=0 then u is constant . You can check "Complex Analysis", by J.Bak-D.J.Newman, chapter 1, however if you need more information, let me now.