Cfrgtkky

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!

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 $|z-1|>4$.

Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}

begin{align}&-frac{2}{z-2}=-frac{2}{(z-1)-1}=\&-frac{2}{(z-1)left(1-frac{1}{z-1}right)}=-2sum_{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.