I've been thinking how to prove that an analytic function $f$ is a constant if the absolute value of $f$ is a constant, but I haven't figured it out yet.
What I was thinking is to use Cauchy-Riemann equations, but it didn't work well...
If this is not true, I would like to know the counterexample...
Here is what I tried:
$|f|=|u+iv|=\sqrt {u^2+v^2}$
Thus $u^2+v^2$ is a constant.
(1) $\displaystyle u\frac {\delta u}{\delta x}+v\frac {\delta v}{\delta x}=0 $
(2) $\displaystyle u\frac {\delta u}{\delta y}+v\frac {\delta v}{\delta y}=0 $
Plug Cauchy Riemann into (2).
$\displaystyle -u\frac {\delta v}{\delta x}+v\frac {\delta u}{\delta x}=0 $
and I'm stuck here...