coefficients of Laurent series


Suppose that f is analytic in the annulus   {z∈ℂ⁢⋮⁢R1<|z-a|<R2},  where R1 may be 0 and R2 may be ∞.  Then the coefficients of the Laurent seriesMathworldPlanetmath (http://planetmath.org/LaurentSeries)

∑n=-∞∞cn⁢(z-a)n

of f can be obtained from

cn=12⁢π⁢i∮γf⁢(t)(t-a)n+1dt (n=0,±1,±2,…), (1)

where the path (http://planetmath.org/ContourIntegral) γ goes anticlockwise once around the point  z=a  within the annulus.  Especially, the residueDlmfPlanetmath of f in the point a is

c-1=12⁢π⁢i⁢∮γf⁢(t)⁢𝑑t. (2)

Remark.  Usually, the Laurent series of a functionMathworldPlanetmath, i.e. the coefficients cn, are not determined by using the integral formula (1), but directly from known series .  Often it is sufficient to know the value of c-1 or the residue, which is used to compute integrals (see the Cauchy residue theorem —  cf. (2)).  There is also the usable

Rule.  In the case that the limit   limz→a⁡(z-a)⁢f⁢(z)  exists and has a non-zero value r, the point  z=a  is a pole of the 1 for the function f and

Res⁡(f;a)=r.

Examples

  1. 1.

    Let  f⁢(z):=1sin⁡z,  and  a=0.  Using the Taylor seriesMathworldPlanetmath of the complex sine we obtain

    limz→0⁡z⁢1sin⁡z=limz→0⁡11-z23!+-…= 1,

    whence  Res⁡(1sin⁡z; 0)=1.  Thus we can write

    ∮γd⁢zsin⁡z= 2⁢π⁢i,

    where the must be chosen such that it encloses only the pole 0 of 1sin⁡z.

  2. 2.

    The Taylor series of the complex exponential function gives the Laurent series

    e1z≡ 1+1z+12!⁢z2+13!⁢z3+…

    which shows that  Res⁡(e1z; 0)=1.

Title coefficients of Laurent series
Canonical name CoefficientsOfLaurentSeries
Date of creation 2013-03-22 15:19:22
Last modified on 2013-03-22 15:19:22
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 15
Author pahio (2872)
Entry type Result
Classification msc 30B10
Related topic LaurentSeries
Related topic TechniqueForComputingResidues
Related topic UniquenessOfLaurentExpansion