How to find the Laurent series of $\frac{1}{z^4(1-z)^2}$ for |z|>1?

Question

A hint is given that $$\frac{1}{(1-\frac{1}{z})^2} = \frac{z^2}{(1-z)^2}$$

and we know that $$\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n$$ for $|w|<1$.

I don't know how to make $\frac{1}{z^4(1-z)^2}$ as $\frac{1}{1-w}$.

Answer 1

You are allowed to differentiate the geometric series termwise, $$ \frac{1}{1-w} = \sum_{n=0}^{\infty} w^n, \quad |w|<1,\tag1 $$ obtaining $$ \frac{1}{(1-w)^2} = \sum_{n=1}^{\infty}n w^{n-1}, \quad |w|<1.\tag2 $$

1. A Laurent series in the set $0<|z|<1$. We apply $(2)$ with $w:=z$, we divide by $z^4$, we make a change of index, obtaining

   $$ \frac{1}{z^4(1-z)^2}=\frac1{z^4}\sum_{n=1}^{\infty}n z^{n-1}=\frac1{z^4}+\frac2{z^3}+\frac3{z^2}+\frac4z+\sum_{n=0}^{\infty}(n+5) z^{n}, $$

   as the sought Laurent series expansion.

2. A Laurent series in the set $|z|>1$. We apply $(2)$ with $w:=\dfrac1z$, we use the given hint, obtaining

   $$ \frac{1}{z^4(1-z)^2}=\frac{1}{z^6(1-\frac{1}{z})^2} =\frac1{z^6}\sum_{n=1}^{\infty}\frac{n}{z^{n-1}}=\sum_{n=1}^{\infty}\frac{n}{z^{n+5}}=\sum_{n=6}^{\infty}\frac{n-5}{z^{n}}, $$

   as the sought Laurent series expansion.