Problem: Show that if $f(z)$ is a nonconstant analytic function on a domain $D$, then $f$ is an open mapping. "Proof": Let $U \subseteq D$ is open, and take $w_0 f(U)$. Then there exists $z_0 \in U$ such that $f(z_0) = w_0$. If $f'(z_0) \neq 0$, then I can invoke the inverse function theorem for analytic functions to show that $w_0$ has an open nhbd contained in $f(U)$, so $f(U)$ is open....For the case of $f'(z_0) =0$, since $f'$ is analytic, it has isolated zeros, meaning I can find $r > 0$ such that $B(z_0,r) \subseteq U$ contains no zeros of $f'$ other than $z_0$. Is it possible to find…