# 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, $\frac{1}{1+z^2}$=$\frac{1}{(z+i)(z-i)}$=$\frac{\frac{-1}{2i}}{z+i} +\frac{\frac{1}{2i}}{z-i}$?

• Good start! The next step depends upon which annulus you want the Laurent expansion for. Apr 11, 2013 at 22:51
• @CameronBuie Thanks, the question stated centered at $z=i$, but I don't recall a region being specified. What can I do in that case?
– Alti
Apr 11, 2013 at 22:54
• Well, you have to make sure that whatever the annulus it won't contain the other pole, for example...I'd propose the annulus $\,0<|z-i|<1\,$ Apr 11, 2013 at 22:58
• @Don: May as well go for $0<|z-i|<2$, right? Apr 11, 2013 at 23:07
• I think he may, @CameronBuie...and it's better imo, since we get a Laurent series in a bigger domain. Apr 11, 2013 at 23:09

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.

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

$$|z-i|<1\implies\frac{1}{z+i}=\frac{1}{z-i+2i}=\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}=\frac{1}{2i}\left(1-\frac{z-i}{2i}+\frac{(z-i)^2}{-4\cdot 2!}+\ldots\right)\implies$$

$$\frac{1}{z^2+1}=\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{2i}\left(1-\frac{z-i}{2i}-\frac{(z-i)^2}{8}+\ldots\right)\right)=$$

$$\frac{1}{2i(z-i)}+\frac{1}{4}-\frac{z-i}{8i}-\frac{(z-i)^2}{32}+\ldots$$