# 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...

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Cauchy-Riemann equations DO work extremely well here... You might wish to expand on your try. – Did Nov 25 '12 at 15:56
C-R in disguise: if $|f|=1$ then $f(z)^{-1} = \overline{f(z)}$. The left hand side is holomorphic and the right hand side can only be holomorphic if $f' = 0$. – WimC Nov 25 '12 at 16:06
@did I added what I did to my question, so could you point out what's wrong with that or how to proceed from that? – Tengu Nov 25 '12 at 16:08
@WimC Thank you for your answer! I finally can prove this :) – Tengu Nov 25 '12 at 16:15
Next: post your own complete answer. – Did Nov 25 '12 at 16:37

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.

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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.

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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.

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If $f$ is analytic on all of $\mathbb C$ then $f$ is constant by Liouville's theorem. However, Chris' argument works in greater generality.

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You probably want to be careful you're not using this result or one very much like it to deduce Liouville... – Ben Millwood Sep 5 '13 at 3:05