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.

Title	Laurent expansion of rational function
Canonical name	LaurentExpansionOfRationalFunction
Date of creation	2013-03-11 19:16:06
Last modified on	2013-03-11 19:16:06
Owner	pahio (2872)
Last modified by	(0)
Numerical id	5
Author	pahio (0)
Entry type	Example
Classification	msc 30B10
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