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.

All the isolated zeros of $f$ in $\Omega$ will be the local minimum of $f$. If the set of zeros of $f$ has a limit point in $\Omega$ then it is identically zero, being holomorphic. If the value of $f$ at a point $z_0$ is non zero then we can use the open set property to prove that it is not a point of minimum value.
