In this question the following statement was shown:
Let $f\colon M\to\mathbb{C}$ be a holomorphic, $M\subseteq\mathbb{C}$ open and connected, $z_0 \in M$ and $f$ is constant in an neighborhood of $z_0$. Then $f$ is constant on the whole $M$.
I wonder if it holds that
If $f\colon M\to\mathbb{C}$ is holomorphic, $M\subseteq\mathbb{C}$ open and connected, $z_0 \in M$ and $|f|$ is constant in an neighborhood of $z_0$. Then $f$ is constant on the whole $M$.
Any help is hugely appreciated.
If we have $|f|=c$ on some $D(z_0,r) \subset M,$ then $f(D(z_0,r))$ is contained in the circle $\{|w|=c\}.$ Thus $f(D(z_0,r))$ is not an open subset of $\mathbb C.$ Hence on $D(z_0,r),$ $f$ is constant by the open mapping theorem. So by the result you first cited, $f$ is constant in $M.$
Yes, if $f$ is holomorphic and $|f|$ is constant on some open set $U$, then $f$ is constant. This follows from writing $f(z)= |f(z)| e^{i F(z)}$ where $F(z) : U \to \mathbb R$.
In this situation note that if $|f(z)|=0$, then $f$ is constant. If it is not zero then, since $\ln$ is analytic, $i\, F(z)$ must be holomorphic on $U$.
But since $F$ maps onto the reals, the Cauchy Riemann equations give you:
$$\partial_x F = \partial_x \Re(F) = \partial_y \Im(F) = 0\\ \partial_y F = \partial_y \Re(F) = -\partial_x \Im(F) = 0$$
This can only be satisfied if $F$ is also constant.
