Laurent expansion of rational function


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

f(z):=2z1+z2

by the powers of z-i.

We first have the partial fraction decomposition

f(z)=1z-i+1z+i (1)

whence the principal part of the Laurent expansion contains 1z-i.  Taking into account the poles  z=±i  of f we see that there are two possible annuli for the Laurent expansion:

a)  The annulus  0<|z-i|< 2.  We can write

1z+i=12i+(z-i)=12i11-(-z-i2i)=12i-z-i(2i)2+(z-i)2(2i)3-+

Thus

2z1+z2=1z-i-n=0(i2)n+1(z-i)n   (0<|z-i|< 2).

b)  The annulus  2<|z-i|<.  Now we write

1z+i=1(z-i)+2i=1z-i11-(-2iz-i)=1z-i-2i(z-i)2+(2i)2(z-i)3-+

Accordingly

2z1+z2=2z-i+n=2(-2i)n-1(z-i)n   (2<|z-i|<).

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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