local minimum of $|f|$ Suppose $f \in H(\Omega)$, where $\Omega\subset\mathbb C$ is an open set. Under what condition can $|f|$ have a local minimum?
Here $|f| = u^2 +v^2 = g$ say. We assumed $f(x,y)= u(x,y) +i v(x,y)$.
Then $g$ has local minimum if $g_{xx} > 0$ and $g_{xx}= 2[u_x^2 +uu_{xx} +v_x ^2 +vv_{xx}]$. So as square terms are positive always, the required condition is $uu_{xx}+vv_{xx} >0$.
I am asking if this a correct answer; if not then please guide me in the right way.
Thanks in advance.

If $f$ has a zero on $\Omega$, then clearly $|f|$ has a local minimum at those points. Otherwise, the open mapping principle prevents $|f|$ from having a local minimum. (If $a \in \Omega$ then $f$ maps open discs centered at $a$ to open sets. In particular if $f(a) \neq 0$ then there are nearby points $z$ where $|f(z)| < |f(a)|$.)
