$\displaystyle f(z) = \sum_{n=0}^{\infty}\frac{a_n}{z^n}$ for the function $\displaystyle \frac{z-1}{z+1}$

so i have that $\displaystyle \frac{z-1}{z+1} = 1 - \frac{2}{z+1}$

does that mean i expand the function at $\displaystyle z=-1$? or $\displaystyle z=\infty$ because the sum is to $\displaystyle \infty$