Laurent expansion of rational function

The Laurent series expansion of a rational function may often be determined using the uniqueness of Laurent series coefficients in an annulus and applying geometric series.  We will determine the expansion of

$$f(z):=\frac{2z}{1+z^2}$$

by the powers of $z-i$.

We first have the partial fraction decomposition

$f(z)={\displaystyle \frac{1}{z-i}}+{\displaystyle \frac{1}{z+i}}$(1)

whence the principal part of the Laurent expansion contains ${\displaystyle \frac{1}{z-i}}$.  Taking into account the poles  $z=\pm i$  of $f$ we see that there are two possible annuli for the Laurent expansion:

a)  The annulus  .  We can write

$$\frac{1}{z+i}=\frac{1}{2i+(z-i)}=\frac{1}{2i}\cdot \frac{1}{1-\left(-\frac{z-i}{2i}\right)}=\frac{1}{2i}-\frac{z-i}{{(2i)}^2}+\frac{{(z-i)}^2}{{(2i)}^3}-+\mathrm{\dots }$$

Thus

b)  The annulus  .  Now we write

$$\frac{1}{z+i}=\frac{1}{(z-i)+2i}=\frac{1}{z-i}\cdot \frac{1}{1-\left(-\frac{2i}{z-i}\right)}=\frac{1}{z-i}-\frac{2i}{{(z-i)}^2}+\frac{{(2i)}^2}{{(z-i)}^3}-+\mathrm{\dots }$$

Accordingly

This latter Laurent expansion consists of negative powers only, but  $z=i$  isn’t an essential singularity of $f$, though.