I've seen this stated all over the internet, but I couldn't find a proof anywhere; also, I've seen a lot of conflicting definitions, but I'm going with the one I learned in class.
If $\mathscr{D} \subseteq \mathbb{C}$ is a domain, and $f: \mathscr{D} \to \mathbb{C}$ is conformal if it is holomorphic and if $f'(z) \neq 0$ for all $z \in \mathscr{D}$.
I was able to prove that conformal mappings perserve angles between curves, and by this I mean that they also perserve angle orientations, not just their measures.
So, strictly speaking, if $f: \mathscr{D} \to \mathbb{C}$ is a conformal mapping, and $\gamma_{1}, \gamma_{2}: [-1, 1] \to \mathscr{D}$ are two smooth curves with $z_{0} = \gamma_{1}(0) = \gamma_{2}(0)$, $\gamma_{1}'(0), \gamma_{2}'(0) \neq 0$, $\gamma_{1}'(0) = \rho_{1}e^{i\varphi_{1}}$, $\gamma_{2}'(0) = \rho_{2}e^{i\varphi_{2}}$, then I define the angle between $\gamma_{1}$ and $\gamma_{2}$ (the order matters) to be $\varphi_{2} - \varphi_{1}$.
What I want to prove is:
If $\mathscr{D} \subseteq \mathbb{C}$ is a domain, and $f: \mathscr{D} \to \mathbb{C}$ perserves angles between smooth curves in the aforementioned sense, prove that $f$ is holomorphic with non-zero derivative.
How do I start?
P.S. I'm aware that the more "natural" definition of conformality would be "angle-perserving", but either way, this part of the equivalence is something that would bother me with either definition.