I am familiar with the definition that a holomorphic function is a conformal mapping of some domain if it has non-zero derivative on that domain (as this is an equivalence with angle-preserving).
However, working through some books there seems to be an equivalence between this definition of conformal map and a holomorphic univalent function.
Can someone please shed some insight on this for me?
Is it necessary that a holomorphic univalent function must have non-zero derivative? Or do we need it to also be surjective?
I don't think that the non-zero derivative definition implies univalence, but again please correct me if I am wrong.
Thanks!
Taking definition that $f:\Omega \to \mathbb{C}$ is conformal if $f$ is holomorphic on $\Omega$ with the property $f'(z) \neq 0$ for all $z \in \Omega$ (where $\Omega$ is an open subset of the complex plane), we have the following: $$f\text{ is univalent} \implies f \text{ is conformal} $$
$\textbf{Proof} $
Let $f$ be a univalent function on $\Omega$. Thus, $f$ is holomorphic and so we can consider the power series expansion$$f(z) = f(z_0) + f'(z_0)(z-z_0)+\dots$$ locally for some $z_0 \in \Omega$. It is easy now to see that this injectivity implies $f'(z_0)\neq 0$, as if we had a vanishing derivative then we would have $f(z)=f(z_0)$ for distinct $z, z_0 \in \Omega$. Since $z_0 \in \Omega$ was arbitrary, we see univalence implies conformality across $\Omega$.