I have a question on biholomorphic functions. I saw the following theorem in lecture:
Let $U,V\subseteq \mathbb{C}$ open subsetes and $f:U\to V$ holomorphic, bijective and $f'(z)\neq 0$ for all $z\in U$. Then $f$ is biholomorphic and it is $$(f^{-1})'=\frac{1}{f'(f^{-1}(w))}$$ for all $w\in V$.
My question is: I thought that holomorphic+bijective is enough to say that $f$ is biholomorphic. Do I need the condition $f'(z)\neq 0$ for all $z\in U$ that the inverse map is holomorphic too? I first thought if a function is holomorphic and bijective, then it follows automatically, that the inverse map is holomorphic.
Best Answer
The assumption $f'(z) \neq 0$ is indeed superfluous. If $f'(a) = 0$, then locally near $a$, $f(z) = z^k\,g(z)$ for some holomorphic $g$ with $g(a) \neq 0$ and $k > 1$ which implies that $f$ is locally $k$-to-$1$ (in particular, not bijective).