Deriving the Laurent series

Question

How do I derive the Laurent series

$$\frac{1}{z^2}=\sum_{n=2}^{\infty} \frac{(-1)^n(n-1)}{(z-1)^n}$$

from $$\frac{1}{(1-z)^2}=\sum_{n=0}^{\infty}(n+1)z^n$$

It looks like I can do some sort of substitution $$z'=\frac{1}{z-1}$$

however I cannot simply to the result.

Any hint would be appreciated

Answer 1

With $\left\vert z - 1\right\vert < 1$: $$ {1 \over z^{2}} = {1 \over \left[1 - \left(1 - z\right)\right]^{2}} = \left.{{\rm d} \over {\rm d}z'}\left(1 \over 1 - z'\right) \right\vert_{z'\ =\ 1 - z} = \left.{{\rm d} \over {\rm d}z'}\sum_{n = 0}^{\infty}z'^{n} \right\vert_{z'\ =\ 1 - z} $$