So I'm trying to find local coordinates to compute the local normal form of a holomorphic function.
I have $f : \mathbb{P}^1 \to \mathbb{P}^1$ given by $f(z) = \frac{z}{(z-1)^2}$.
Now we have a nice coordinate system $\varphi(z)= \frac{1}{z-1}$ and $ \psi(z) = \frac{-1 + \sqrt{1+4z}}{2}$ around $\infty$ and $f(\infty) = 0$ respectively.
So we have the lovely fact that $\psi \circ f \circ \varphi^{-1}(z) = z$. Now I believe this is the local normal form of $f$ around $\infty$. This should also work nicely with only some mild alterations to $\varphi$ for most points in the domain. With I believe the only exception of $z = 1 \in \mathbb{P}^1$.
I think that the normal form around $z=1$ should have power 2, since there is a pole of that order there. I can't for the life of me figure it out though. Any ideas?
Around $z = 1$, you want a normal form of $w \mapsto w^2$. Since $f$ has a pole in $1$, you can choose the chart $\psi(z) = \frac1z$ around $\infty$, so you want
$$w^2 = \frac{(z-1)^2}{z},$$
with a little abuse of notation. Thus you can use the chart
$$\varphi(z) = \frac{z-1}{\sqrt{z}}$$
around $1$. I suggest using the branch of the square root that has $\sqrt{1} = 1$. A small computation shows
$$\varphi^{-1}(w) = \left(\sqrt{1+ \frac{w^2}{4}} + \frac{w}{2}\right)^2,$$
and as desired
$$\psi \circ f \circ \varphi^{-1} \colon w \mapsto w^2.$$