Laurent series of $\frac{1}{z^4+z^2}$ about $z=0$ Okay, I get $\frac{1}{z^4+z^2}=\frac{1}{z^2}-\frac{1}{z^2+1}$
And I know I need to use geometric series sum but couldn't quite find it. Can someone give me a hint?

One of the two terms is easy $$\frac{1}{1+z^2} = \frac{1}{1-(-z^2)}=\sum_{m=0}^{+\infty} (-1)^mz^{2m}.$$ For the other, notice that $\frac{1}{z^2}$ is already in a Laurent form at $z=0.$ So the developpement is $$\frac{1}{z^4+z^2} = \frac{1}{z^2} - \sum_{m=0}^{+\infty} (-1)^mz^{2m}.$$
