<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] Use the fact that\begin{align}\frac{-2}{z-2}&=\frac{-2}{-1+(z-1)}\\&=\frac2{1-(z-1)}\\&=-2\sum_{n=-\infty}^{-1}(z-1)^n\end{align}and that\begin{align}\frac3{z+3}&=\frac3{4+(z-1)}\\&=\frac34\times\frac1{1+\frac{z-1}4}\\&=-\frac34\sum_{n=-\infty}^{-1}\frac{(-1)^n}{4^n}(z-1)^n.\end{align}
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