<s>[INST] Laurent series for $\frac1{z^2+1}$ I have this problem:
Find the Laurent series around $z=0$, for $\dfrac{10}{(z+2)(z^2+1)}$ in the region $1<|z|<2$.
I did partial fractions and found this: $\dfrac{2}{z+2}-\dfrac{2z-4}{z^2+1}$,
then I have to know what's the Laurent series for $\dfrac{1}{z^2+1}$ to solve the problem.
Do you know that series? Thank you.
 [/INST] Expanding geometrically,
$$\frac{2}{z+2} = \frac{1}{ 1 + \frac z 2} = \sum_{k=0}^\infty (-1)^k \left( \frac z 2\right)^k $$
Similarly, 
$$\frac{2z-4}{z^2 + 1} = \frac{2z-4}{z^2}\cdot \frac 1 {1 + \frac{1}{z^2}} = \frac{2z-4}{z^2} \sum_{k=0}^{\infty}(-1)^k \left( \frac 1 {z^2}\right)^k$$
</s>