PlanetMath: coefficients of Laurent series

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (194)
Orphanage
Unclass'd
Unproven (498)
Corrections (25)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

coefficients of Laurent series

(Result)

Suppose that is analytic in the annulus $\{z\in\mathbb{C}\mid\,\, R_1 < \vert z-a\vert < R_2 \}$ , where may be 0 and may be $\infty$ . Then the coefficients of the Laurent series expansion

$\displaystyle \sum_{n = -\infty}^\infty c_n (z-a)^n$

can be obtained from

$\displaystyle c_n = \frac{1}{2\pi i}\oint_{\gamma} \frac{f(t)}{(t-a)^{n+1}}\,dt \quad (n = 0,\,\pm 1,\,\pm 2,\,\ldots),$

(1)

where the path $\gamma$ goes anticlockwise once around the point

within the annulus. Especially, the residue of

in the point

$\displaystyle c_{-1} = \frac{1}{2\pi i}\oint_{\gamma} f(t)\,dt.$

(2)

Remark. Usually, the Laurent series of a function, i.e. the coefficients , are not determined by using the integral formula (1), but directly from known series expansions. 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 $\displaystyle\lim_{z\to a}(z-a)f(z)$ exists and has a non-zero value , the point is a pole of the order 1 for the function and

$\displaystyle \operatorname{Res}(f;\,a) = r.$

Examples

Let $f(z) := \frac{1}{\sin{z}}$ , and . Using the Taylor series of the complex sine we obtain
$\displaystyle \lim_{z\to 0}z\frac{1}{\sin{z}} = \lim_{z\to 0}\frac{1}{1-\frac{z^2}{3!}+-\ldots} = 1,$
whence $\operatorname{Res}(\frac{1}{\sin{z}};\,0) = 1$ . Thus we can write
$\displaystyle \oint_{\gamma}\frac{dz}{\sin{z}} = 2\pi i,$
where the path must be chosen such that it encloses only the pole 0 of $\frac{1}{\sin{z}}$ .
The Taylor series of the complex exponential function gives the Laurent series
$\displaystyle e^{\frac{1}{z}} \equiv 1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\ldots$
which shows that $\operatorname{Res}(e^{\frac{1}{z}};\,0) = 1.$