Laurent Series of $\frac{1}{(1+z^2)^2}$

Question

Expanded to the Laurent Series at the deleted neighbourhood of $$z=i$$ and try to give the Convergence range

I'm trying to make $$\frac{1}{(1+z^2)^2}=\bigg(\frac{-1}{2z}\bigg)\bigg(\frac{1}{1+ z^2 }\bigg)\text {'}$$

$$\frac{1}{1+ z^2}={1-z^2+z^4-z^6...}=\sum_{n=0}^{\infty} {(-1)^ n(z^{2n})}$$

$$\bigg(\frac{1}{1+ z^2 }\bigg)\text {'}= \sum_{n=0}^{\infty} {(2n)(-1)^ n(z^{2n-1})}$$

so that

$$\frac{1}{(1+z^2)^2}=\bigg(\frac{-1}{2z}\bigg)\sum_{n=0}^{\infty} {(2n)(-1)^ n(z^{2n-1})}$$

$$=-\sum_{n=0}^{\infty} {(n)(-1)^n(z^{2n-2})}$$ But it seems not right...

Answer 1

I really can't follow what you are doing. You start by saying, "I'm trying to make $$\frac{1}{(1+z^2)^2}=\bigg(\frac{-1}{2z}\bigg)\bigg(\frac{1}{1+ z^2 }\bigg),$$ but you surely know that this isn't a true statement. I also don't see what it has to do with what follows.

Let me get you started: $$\frac{1}{(z^2+1)^2}=\frac{1}{(z+i)^2(z-i)^2} $$ Now $1/(z+i)^2$ is analytic at $z=i$ and so can be developed in a Taylor series about $z=i$. Once you do this, you can simply multiply the series by $1/(z-i)^2$ to get the desired Laurent series.

Answer 2

Hint: Series for $\frac{1}{1+z^2}$ is $\frac{1}{1-(-z^2)}=\sum_{k=0}^{\infty}(-z^2)^k$.