Laurent series of $\frac{z}{z^2+1}$

Question

I am interested in the Laurent series of $f(z)=\frac{z}{z^2+1}$ in $D_{0,2}(i)$.

What I did:

We have $\displaystyle f(z)=\frac{1}{2}\frac{1}{z+i}+\frac{1}{2}\frac{1}{z-i}$, so

1. $\displaystyle \frac{1}{2}\frac{1}{z+i}=\frac{1}{2i}\frac{1}{1-\left(-\frac{z}{i}\right)}=\frac{1}{2i}\sum_{n=0}^\infty(iz)^n$

2. $\displaystyle \frac{1}{2}\frac{1}{z-i}=-\frac{1}{2i}\frac{1}{1-\left(\frac{z}{i}\right)}=-\frac{1}{2i}\sum_{n=0}^\infty(-iz)^n$

My problem is that these series only converge in $D_{0,1}(0)$ so how do I obtain the laurent Series centered at $i$?

Answer 1

It is the right idea to use the geometric series. Try it this way: $$ \frac1{z+i}=\frac1{2i+z-i}=\frac1{2i}\cdot\frac1{1+\frac{z-i}{2i}}=\frac1{2i}\cdot\frac1{1-\left(-\frac{z-i}{2i}\right)}=\frac1{2i}\sum_{n=0}^\infty\left(\frac{i}2\right)^n(z-i)^n. $$ The term $\frac1{z-i}=(z-i)^{-1}$ is already a part of the Laurent series and stays as it is. Together you get $$ f(z)=\frac12(z-i)^{-1}+\sum_{n=0}^\infty\frac1{4i}\left(\frac{i}2\right)^n(z-i)^n. $$

Answer 2

The point is that a Laurent series centered at $i$ will look like $$\sum_{n>>-\infty}a_n (z-i)^n. $$ In your case, you've done some of the work by writing $$f(z)=\frac{1}{2}\frac{1}{z+i}+\frac{1}{2}\frac{1}{z-i}. $$ Notice that the second term already is in the desired form. With the inspirational example of $$\frac{a}{1-r}=\sum_{n\geq 0}a r^n $$ in mind, let us work towards converting $\frac{1}{z+i}$ into a similar form. Some twisting will yield $$\frac{1}{z+i}= \frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}} $$ and now we can use the above series to expand this.