Laurent expansion of $f(z) = \frac{1}{z^2-1}$ around $0<|z+1|<2$

Question

I am trying to make a Laurent series of $f(z)$ around $0<|z+1|<2$, but I don't know what to do exactly with this interval. I did the following: \begin{align} \frac{1}{z^2-1} = \frac{1}{(z+1)(z-1)} &= \frac{1}{2}\frac{1}{z-1} - \frac{1}{2}\frac{1}{z+1} \\ &= -\frac{1}{2}\sum_{n=0}^{\infty}z^n - \frac{1}{2}\sum_{n=0}^{\infty}(-1)^nz^n \end{align} But this holds only for $|z|<1$. How do I adapt to the interval?

Answer 1

You expanded the function in the domain $|z|<1$ which is NOT what the exercise is asking. Instead, let $w=z+1$, then $$f(z) = \frac{1}{z^2-1}=\frac{1}{w(w-2)}=\frac{1}{-2w}\cdot\frac{1}{1-w/2}$$ Now expand the RHS with respect to $w$ in the domain $0<|w|/2<1$ and finally go back to the variable $$.