Laurent series $z=i$ To find the Laurent series expansion for $\frac{1}{1+z^2}$, centered at $z=i$ would using partial fraction decomposition be the right idea? So, $\frac{1}{1+z^2}$=$\frac{1}{(z+i)(z-i)}$=$\frac{\frac{-1}{2i}}{z+i} +\frac{\frac{1}{2i}}{z-i}$?
 A: Note that for $z\neq\pm i$ we can write $$\frac1{z+i}=\cfrac1{z-i+2i}=\frac1{2i}\cdot\cfrac1{1-\left(-\frac{z-i}{2i}\right)}$$ and $$\frac1{z+i}=\cfrac1{z-i+2i}=\frac1{z-i}\cdot\cfrac1{1-\left(-\frac{2i}{z-i}\right)}.$$
Now, one of these can be expanded as a multiple of a geometric series in the annulus $|z-i|>2$ and the other can be expanded as a multiple of a geometric series in the annulus $0<|z-i|<2$. That is, we will use the fact that $$\frac1{1-w}=\sum_{k=0}^\infty w^k$$ whenever $|w|<1$. You should figure out which annulus works for which rewritten version, and find the respective expansions in both cases. That will give you two different Laurent expansions of your function, valid in two different annuli.
A: First:
$$\frac{1}{z^2+1}=\frac{1}{(z-i)(z+i)}=\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)$$
Now
$$|z-i|<1\implies\frac{1}{z+i}=\frac{1}{z-i+2i}=\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}=\frac{1}{2i}\left(1-\frac{z-i}{2i}+\frac{(z-i)^2}{-4\cdot 2!}+\ldots\right)\implies$$
$$\frac{1}{z^2+1}=\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{2i}\left(1-\frac{z-i}{2i}-\frac{(z-i)^2}{8}+\ldots\right)\right)=$$
$$\frac{1}{2i(z-i)}+\frac{1}{4}-\frac{z-i}{8i}-\frac{(z-i)^2}{32}+\ldots$$
