Find the Laurent series for $f(z) = \frac{z-12}{z^2 + z - 6}$ valid for $(a) \ \ \ 1 < |z-1| <4$
$(b) \ \ \ |z-1| > 1$
$(c) \ \ \ |z-1| < 4$
We have $$f(z)= f(z) = \frac{z-12}{z^2 + z - 6} = \frac{-2}{z-2} + \frac{3}{z+3}$$
I know the answer for $(a)$
$$f(z) = -2 \sum_{n=0}^{\infty} \frac{1}{(z-1)^{n+1}} + 3 \sum_{n =0}^{\infty }(-1)^{n}\frac{(z-1)^{n}}{4^{n+1}}$$
How Can I find the Laurent series for $(b)$ and $(c)$ ?
Thank you ...