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.

Realize the last series you ask about as a geometric series...
$$
\frac{1}{z^2+1} = \frac{1}{z^2}\;\frac{1}{1-(-1/z^2)} =
\frac{1}{z^2} - \frac{1}{z^4} +\frac{1}{z^6} - \frac{1}{z^8}+\dots
$$
