Let $f: B \to \mathbb{C} $ be analytic and $f'(z) \neq 0 $ for all $z \in B $. Then $f$ maps open sets in $B$ to open sets.
This is problem is supposed to be solved using basic complex analysis. I was thinking on using the inverse function theorem:
For the inverse function theorem to be applied we must have that $f'(z)$ must be continuous. Can we assume $f'$ is continuos? IF so, then we can find neighborhood $N$ of some $z_o \in B$ and $V$ of $f(z_0)$ such that $f: N \to V $. since we can apply this to every point on $B$, then is it valid to conlude that $f$ maps open set to open sets?
Best Answer
Yes you can apply the inverse function theorem. There are however other methods.
$f^\prime$ is indeed continuous. Remember that an analytic function can be expanded locally to a power series. And a power series is indefinitely complex differentiable.