Let $f(z)$ be a holomorphic function defined on a punctured neighborhood of $z=0$ with non-essential singularity of degree $d$ at $0$ (namely, $f(z)=z^dh(z)$, where $h(z)$ is a holomorphic function defined around $0$ with $h(0)\neq0$). I presume the following two interrelated facts are well known:

Proposition 1.Given $k\in\mathbb{Z}_+$, there is a conformal local coordinate $w=w(z)$ centered at $0$ which brings the meromorphic $k$-differential $f(z)dz^k$ into the following normal form: $$ f(z)dz^k=\left\{ \begin{array} {}w^ddw^k&\text{ if }d\notin\{-k,-2k,\cdots\}\\ Rw^{-k}dw^k\quad (R\in\mathbb{C}^*)&\text{ if }d=-k\\ w^{-k}(w^{-l}+A)^kdw^k\quad (A\in\mathbb{C})&\text{ if }d=-(l+1)k\text{ with }l\in\mathbb{Z}_+ \end{array} \right. $$ Moreover, $R$ and $A^k$ are invariants of the meromorphic $k$-differential at the singularity, not depending on the choice of $w$.

Proposition 2.Given $\alpha\in(0,+\infty)$, there is conformal local coordinate $w=w(z)$ centered at $z=0$ which brings the flat Riemannian metric $|f(z)|^{2\alpha}|dz|^2$ with singularity at $0$ into the normal form $$ |f(z)|^{2\alpha}|dz|^2=\left\{ \begin{array} {}|w|^{2\alpha d}|dw|^2 &\text{ if }\alpha d \notin\{-1,-2,\cdots\}\\ R|w|^{-2}|dw|^2\quad(R>0)&\text{ if }\alpha d=-1\\ |w|^{-2}|w^{-l}+A|^{2}|dw|^2\quad(A\geq0)&\text{ if }\alpha d=-(l+1) \text{ with }l\in\mathbb{Z}_+ \end{array} \right. $$ Moreover, $R$ and $A$ are invariants of the flat metric at the singularity, not depending on $w$.

However, the only literature addressing exactly these issues that I can find is the book "Quadratic differentials" by Strebel, where a proof of Proposition 1 is given only in the $k=2$ case. In the mathoverflow discussion Analog of residue for meromorphic quadratic differentials the answer of Robert Bryant is exactly Proposition 1 but no reference is given.

Question:Does there exist a literature containing a proof of Proposition 1 for general $k$ or Proposition 2?

Surfaces of mean curvature one in hyperbolic space, Astérisque 154–155 (1987), 341–347. See Proposition 4, if you are curious. $\endgroup$ – Robert Bryant Jun 7 '18 at 13:10