# Theorem about constant complex functions

Theorem: Let $$f$$ be analytical on $$\Omega \subset \mathbb C$$ and let $$|f(z)|=k$$ be a constant in $$\Omega$$. Then it follows that $$f$$ is constant.

Question: My book now says "it's clear that the theorem is wrong if $$f$$ is a complex function of real variables... But that kind of confuses me. Let's pick $$f(z)=1$$. Then $$f$$ is analytical on $$\mathbb C$$, especially on $$\Omega$$. Also we have $$|f(z)|=1,\quad \forall z \in \Omega$$ but $$f$$ is a complex function of a real variable, and apparently, that shouldn't work.

Can maybe elaborate someone what exactly is meant with that theorem?

• The text means $f\colon \Bbb R \to \Bbb C$. E.G. $f(x) = \exp(\mathrm i 2\pi x), x \in \mathbb R$. Then $f(x) = \cos(2\pi x) + \mathrm i \sin(2\pi x)$ and $\cos, \sin$ are real analytic. But $|f|=1$ while $f$ is non-constant. – xbh Nov 12 '18 at 14:16
• For this theorem, try C-R equations. – xbh Nov 12 '18 at 14:17

If $$f(z)=1$$ for $$z \in \Omega$$, then $$f$$ is a complex function of a complex variable !

Look at $$g(t)=e^{it}$$ for real $$t$$. $$g$$ is differentiable on $$\mathbb R$$, we have $$|g(t)|=1$$ for all $$t$$, but $$g$$ is not constant.

The statement in your book just means that there are some complex functions of real variables for which the Theorem does not hold - it does not mean that the Theorem fails to hold for all such functions.

One example for which the Theorem fails to hold is

$$f(x) = \cos(x) + i \sin(x)$$

for which $$|f(x)|=1 \space \forall x \in \mathbb{R}$$, but clearly $$f$$ is not a constant function.

• So f is not analytical? – xotix Nov 12 '18 at 16:08
• Not that. $f$ is real analytic, and the norm $|f|$ is constant, but itself is not a constant function. This explains why "the theorem is wrong if $f$ is a complex function of real variables". – xbh Nov 12 '18 at 16:40