I have $\dfrac{z}{(z-1)(z+1)}$ and wish to find the Laurent expansion about $z = 1$. I have simplified this down to $\dfrac{z}{z^2 - 1}$ but I'm unsure of where to go from here. A hint has been given that I can use a substitution but I don't know what to use.
Any help at all is greatly appreciated.
Here how you advance
$$= \dfrac{z}{(z-1)(z+1)} = \dfrac{(z+1)-1}{(z-1)(z+1)}= \frac{1}{z-1}- \dfrac{1}{(z-1)(z+1)} $$
$$=\frac{1}{z-1}- \dfrac{1}{(z-1)(2+(z-1))} $$
$$=\frac{1}{z-1}-\dfrac{1}{2(z-1)\left(1+\frac{(z-1)}{2}\right)} $$
$$=\frac{1}{z-1}-\dfrac{1}{2(z-1)}\sum_{k=0}^{\infty} (-1)^k\frac{(z-1)^k}{2^k}=\dots. $$