Let $B$ be a finitely connected bounded planar domain, and suppose $z_0$ is a boundary point of $B$ for which there is an arc $\gamma : [0,1) \rightarrow B $ (injective continuous function) satisfying $\lim_{t \rightarrow 1} \gamma(t) = z_0$, and in the connected component of $U(z_0, r) \cap B$ containing the tail part of $\gamma$, say $S$, some holomorphic bijection (biholomorphism) $f : B \rightarrow f(B)$ , where $f(B)$ is a finitely connected bounded domain, whose boundary consists of disjoint analytic Jordan curves, has the expansion $f(z) = f(z_0) + (z-z_0)^{\frac{1}{\alpha}}(A + o(1)), z \rightarrow z_0$ where $0 < \alpha \leq 2$, where $A \neq 0$, holds in $S$.
Then it is a fact that the quantity $\alpha$ should only depend on $B$, $z_0$, and the neighbourhood $S$, that is, for any other holomorphic bijection onto a finitely connected domain bounded by finitely many analytic Jordan curves, having an expansion of the same form in another such connected open neighbourhood containing the tail part of the curve $\gamma$, the exponent in its expansion will also be $\alpha$.
This is apparently “clear”, but I am certainly not able to see why this needs to be true, I can see that the exponent $\alpha$ is essentially related to the angle at the point $z_0$ in such a neighbourhood, but this is only for nice enough domains $B$, and Im unsure how to tie this in to prove that $\alpha$ is invariant.
Thank you for any suggestions.
$\textbf{Speculation}$: I think this may be because the exponent as defined is multiplicative, and for a conformal map from a finitely connected domain to another, both bounded by analytic curves, the exponent at any boundary point is always exactly $1$ by the Schwarz reflection principle (i.e since such maps extend to be holomorphic in whole neighbourhoods of their boundaries), therefore putting these two results together we may obtain the result. If so I would be grateful for verification this is why the exponent is invariant.