[Math] If $f$ is analytic then $|f|$ is not constant unless $f$ is constant

Question

Let $f$ be analytic in a domain (open connected set of the domain of definition) (that is, every point of that domain is such that $f'$ exists in that point and also in a neighborhood of radius $r>0$ of that point). Show that his absolute value $|f|$ can't be constant unless $f$ is constant too.

Firstly, I'm a little confused about the statement. I must prove $|f|$ is constant $\Rightarrow f$ is constant?

Assuming that's what I must show, here's my attempt: if $f$ and $\overline{f}$ are both analytic then $f$ is constant so if $f$ is not constant and $f$ is analytic we have that $\overline{f}$ is not analytic at some point $p$. But $|f| = \sqrt{f\overline{f}}$ so $|f|$ is also not analytic at the point $p$ so it can't be constant because constant functions are analytics at all points of their domain.

Is my attempt correct? Thanks in advance.

Answer 1

Suppose that $|f|\equiv k$ for some $k\in[0,+\infty)$.

1. If $k=0$, then $f\equiv0$.
2. Otherwise, the image of $f$ is contained in the circle centered at $0$ with radius $k$. Such a set contains no open non-empty subset of $\mathbb C$. But, if $f$ was not constant, then by the open mapping theorem, its image would be an open (and obviously non-empty) subset of $\mathbb C$.