<s>[INST] Laurent series of $ \frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$ 
How do you find the Laurent series for $f(z) = \dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$? 

I know that $f(z) = \dfrac{z-12}{z^2 + z - 6} = \dfrac{-2}{z-2}  + \dfrac{3}{z+3}$
It is easy for me to extract a series for $\dfrac{3}{z+3}$, but have no idea how to do it for $\dfrac{-2}{z-2}$.
Please help? Thank you! 
 [/INST] Hint. Starting from your partial fraction decomposition, we have that
$$\frac{z-12}{z^2 + z - 6} = -\frac{2}{u-1}  + \frac{3}{u+4}=-\frac{2/u}{1-1/u}  + \frac{3/u}{1+4/u}$$
where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$. 
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