If the absolute value of an analytic function $f$ is a constant, must $f$ be a constant? 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...
 A: Since there has been no posted/accepted answer, I'll post my own solution.
Let $f = u + iv$, so $|f| = |u + iv| = \sqrt{u^2 + v^2}$.
This implies $u^2 + v^2 = k$ for some constant $k$. If $k = 0$ then we are done, so consider $ k \ne 0$. Now taking partial derivatives we find
$$uu_x + vv_x = 0$$
$$uu_y + vv_y = 0$$
Using Cauchy-Riemann equations
$$uv_y + vv_x = 0$$
$$-uv_x + vv_y = 0$$
Equating both sides gives $ v_x(v+u) + v_y(u-v) = 0$ and the result follows immediately.
A: This method uses the fact that if $f$ and $\bar{f}$ are both analytic then $f$ is constant. If $|f|=0$ then $f$ is always zero. If $c=|f|>0$ we have $c^2=f\bar{f}$ then $\bar{f}=c^2/f$. Since $f\neq 0$ it follows that $\bar{f}$ is analytic, and hence $f$ is constant.
A: You can also deduce this from the open mapping theorem: a nonconstant holomorphic function is an open map. If $|f|$ is constant, then $f(\mathbf C)$ is contained in the circle of radius $|f|$, which has empty interior. Hence $f$ is constant.
A: If $f$ is analytic on all of $\mathbb C$ then $f$ is constant by Liouville's theorem. However, Chris' argument works in greater generality. 
