Laurent series of $ \frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$

Question

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!

Answer 1

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 $|z-1|>4$.

Answer 2

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}

Answer 3

\begin{align}&-\frac{2}{z-2}=-\frac{2}{(z-1)-1}=\\&-\frac{2}{(z-1)\left(1-\frac{1}{z-1}\right)}=-2\sum_{n=1}^{\infty}\frac{1}{(z-1)^n}\end{align}

That is valid in the region $|z-1|>1$. With similar reasoning you get a series for the other term valid in the region $|z-1|>4$, which is more restrictive than $|z-1|>1$, so it goes for both series expansions.