Laurent series for function $(z-1)/z^2$ in domain $ |z - 1| > 1$

Question

I'm having trouble with the following Laurent series question: Laurent series for function $$\frac{z-1}{z^2}$$ in domain $|z - 1| > 1$.

Specifically, how to modify $z^2$ into powers of $(z-1)$ and then turning into a geometric series.

Answer 1

Hint: use the Taylor series of $(1+u)^{-2}$ at $0$ after you have observed that $$ \frac{z-1}{z^2}=\frac{z-1}{((z-1)+1)^2}=\frac{1}{(z-1)}\cdot\frac{1}{\left(1+\frac{1}{z-1} \right)^2}\quad\mbox{and}\quad \Big| \frac{1}{z-1}\big|<1\;\iff\;|z-1|>1. $$

Answer 2

You can substitute $$y=\frac{1}{z-1}$$ and expand the function in powers of $y$. But if you do this, you obtain $$z=\frac{1}{y}+1$$ and if you plug this into your function, you obtain $$\frac{y}{(y+1)^2}$$ which you can expand normally.