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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?

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    $\begingroup$ 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. $\endgroup$ – xbh Nov 12 '18 at 14:16
  • $\begingroup$ For this theorem, try C-R equations. $\endgroup$ – xbh Nov 12 '18 at 14:17
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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.

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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.

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  • $\begingroup$ So f is not analytical? $\endgroup$ – xotix Nov 12 '18 at 16:08
  • $\begingroup$ 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". $\endgroup$ – xbh Nov 12 '18 at 16:40

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