Question

9)Find Laurent series expansion of the function 1/ z²+z-6 in the regions

1<|z-1 <4 and |z-1|> 4.

Answer 1

From partial fraction

$\frac{1}{(z^2+z-6)}=−\frac{1}{z−1}+\frac{1}{z−2}\\$

so on either annulus, it is enough to find the Laurent series expansions of the functions $\frac{1}{z−1}$ and $\frac{1}{z−2}$ and then combine them term by term.

If $|z|>2$  then note that

$\frac{1}{z−2}=\frac{1}{z} \frac{1}{1−2z^{−1}}$

and the fact that |z|>2

|z|>2 implies that |2z

−1

|<1

|2z−1|<1. If you recall the geometric series formula,