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.
 A: 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 $|f(z)| < |f(a)|$.)
A: 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.
A: If $f$ is a constant, $|f|$ has a local minimum at any point of $\Omega.$
If $f$ is not a constant and there exists $z\in\Omega$ s.t. $f(z)=0,$ then the minimum occurs at those points $z\in\Omega.$
If $f$ is not a constant and there isn't any $z\in\Omega$ s.t. $f(z)=0,$ then $\frac{1}{f}$ doesn't have singularities on $\Omega.$ So, $f\in h(\Omega)$ and $\frac{1}{f}\in h(\Omega).$ Let suppose that $|f|$ occurs minimum at $a\in\Omega.$ Then, by the Principle of Maximum Modulus for $\frac{1}{f}$ we have that $\vert\frac{1}{f(a)}\vert<\{\vert\frac{1}{f(z)}\vert\vert z\in\partial D\}$ for any neighborhood $\bar{D}\subset\Omega$ of $a,$ i.e. $|f(a)|>|f(z)|$ for a $z\in\partial D\subset\Omega$ witch contradicts the supposition that $a$ was a local minimum. This means that under there conditions, |f| does not occur a local minimum on $\Omega.$
