Complex Analysis/Example Computation with Laurent Series

In this learning resource, rational functions are developed into Laurent series to extract the residue.

From a Rational Function to a Laurent Series

Initially, a simple rational function of the following form is given:

• $f:G\to ℂ$ with
• $G:=ℂ\setminus \{-a\},a\in ℂ$
• $f(z):=\frac{1}{a+z}$

The goal is to develop it into a Laurent series with the expansion point $z_o\in ℂ$..

Definition of Constants

The following constants are defined to better illustrate the operations:

• $b:=-a$
• $c:=b-z_o=-a-z_o$
• $c_n:=\frac{1}{(b-z_o{)}^{n+1}}=\frac{1}{(-z_o-a{)}^{n+1}}$
• $q:=\frac{z-z_0}{c}$

Transformation into a Laurent Series

Let $z_o=-a$, then:

$\begin{array}{cc}f(z)& =\frac{1}{a+z}=-\frac{1}{b-z}(b:=-a)\\ & =-\frac{1}{b\underset{=0}{\underbrace{-z_o+z_o}}-z}=-\frac{1}{(b-z_o)-(z-z_o)}\\ & =-\frac{1}{(\underset{=c}{\underbrace{b-z_o}})-(z-z_o)}=-\frac{1}{c-(z-z_o)}\\ & =-\frac{1}{c}\cdot \frac{1}{1-\frac{z-z_o}{c}}=-\frac{1}{c}\cdot \frac{1}{1-\underset{:=q}{\underbrace{\frac{z-z_o}{c}}}}\\ & =-\frac{1}{c}\cdot \frac{1}{1-q}=-\frac{1}{c}\cdot \sum _{n=0}^{+\mathrm{\infty }}{q}^n=-\sum _{n=0}^{+\mathrm{\infty }}\frac{(z-z_o{)}^n}{c^{(n+1)}}\\ & =-\sum _{n=0}^{+\mathrm{\infty }}\frac{(z-z_o{)}^n}{(z_o-a{)}^{n+1}}=-\sum _{n=0}^{+\mathrm{\infty }}\underset{c_n:=}{\underbrace{\frac{1}{(-z_o-a{)}^{n+1}}}}\cdot (z-z_o{)}^n\\ & =-\sum _{n=0}^{+\mathrm{\infty }}{c}_n\cdot (z-z_o{)}^n\end{array}$ :

The residue $re{s}_{z_0}(f)=0$ ,since in the Laurent expansion, the principal part coefficients are all zero (i.e., the principal part vanishes).

Tasks

• Why is the condition required for the above calculation Laurent Series (or power series)$z_o=-a$?
• Compute the Laurent series for $z_o=-a$ and determine the Residue of the Laurent expansion for $f(z):=\frac{1}{a+z}$ in $z_o=-a$ at!***

Factored Powers with Expansion Point in the Denominator

Definition of the Function

First,we are given a simple rational function of the form:

• $g:G\to ℂ$ mit
• $G:=ℂ\setminus \{-a\},a\in ℂ$
• $g(z):=\frac{1}{(z-z_o{)}^m\cdot (a+z)}$

The goal is to develop it into a Laurent series with the expansion point $z_o\in ℂ\setminus \{-a\}$.

Definition of Constants

The following constants are defined to better illustrate the operations:

• $c:=z_o-a$
• $c_n:=\frac{1}{(b+z_o{)}^{n+1}}=\frac{1}{(z_o-a{)}^{n+1}}$
• $q:=\frac{z-z_o}{c}=\frac{z-z_o}{z_o-a}$

Transformation into a Laurent Series

$\begin{array}{cc}g(z)& =\frac{1}{(z-z_o{)}^m\cdot (a+z)}=\frac{1}{(z-z_o{)}^m}\cdot \frac{1}{a+z}(c:=z_o-a)\\ & =-\frac{1}{(z-z_o{)}^m}\cdot \frac{1}{z_o-a}\cdot \frac{1}{1-\frac{z-z_o}{z_o-a}}=-\frac{1}{(z-z_o{)}^m}\cdot \frac{1}{c}\cdot \frac{1}{1-q}\\ & =-\frac{1}{(z-z_o{)}^m}\cdot \frac{1}{c}\cdot \sum _{n=0}^{+\mathrm{\infty }}{q}^n=-\sum _{n=0}^{+\mathrm{\infty }}\frac{(z-z_o{)}^n}{c^{n+1}}\\ & =-\frac{1}{(z-z_o{)}^m}\cdot \sum _{n=0}^{+\mathrm{\infty }}\frac{(z-z_o{)}^n}{(z_o-a{)}^{n+1}}\\ & =-\frac{1}{(z-z_o{)}^m}\cdot \sum _{n=0}^{+\mathrm{\infty }}\underset{c_n:=}{\underbrace{\frac{1}{(z_o-a{)}^{n+1}}}}\cdot (z-z_o{)}^n\\ & =-\frac{1}{(z-z_o{)}^m}\cdot \sum _{n=0}^{+\mathrm{\infty }}{c}_n\cdot (z-z_o{)}^n\\ & =-\sum _{n=-m}^{+\mathrm{\infty }}{c}_{n+m}\cdot (z-z_o{)}^n\end{array}$

the residue $re{s}_{z_0}(g)=c_{-1+m}=\frac{1}{(b-z_o{)}^{-1+m+1}}=\frac{1}{(b-z_o{)}^m}=\frac{1}{(-z_o-a{)}^m}$.

Laurent Series with Infinite Principal Part Terms

A simple rational function of the following form is given:

• $h:G\to ℂ$ with
• $G:=ℂ\setminus \{-a\},a\in ℂ\setminus \{-a\}$
• $h(z):=\frac{(z-z_0{)}^m}{z+a}$

The goal is to develop it into a Laurent series with the expansion point $z_o\in ℂ\setminus \{-a\}$.

Definition of Constants

The following constants are defined for better clarity:

• $b:=-z_0-a$
• $c_n:=b^n$
• $q:=\frac{b}{z-z_o}=-\frac{z_0+a}{z-z_o}$

Transformation into a Laurent Series with

$\begin{array}{cc}h(z)& =\frac{z-z_0}{z+a}=\frac{z-z_0}{z\underset{=0}{\underbrace{-z_o+z_o}}+a}\\ & =\frac{z-z_0}{(z-z_o)-\underset{=b}{\underbrace{(-z_o-a)}}}=\frac{z-z_0}{(z-z_o)-b}\\ & =\frac{1}{1-\frac{b}{z-z_0}}=\frac{1}{1-\underset{:=q}{\underbrace{\frac{b}{z-z_o}}}}\\ & =\sum _{n=0}^{+\mathrm{\infty }}{q}^n=\sum _{n=0}^{+\mathrm{\infty }}\frac{b^n}{(z-z_o{)}^n}\\ & =\sum _{n=0}^{+\mathrm{\infty }}{b}^n\cdot (z-z_o{)}^{-n}=\sum _{n=-\mathrm{\infty }}^0{c}_{-n}\cdot (z-z_o{)}^n\end{array}$

The residue$re{s}_{z_0}(h)=c_1=b^1=-z_o-a$

Transformation into a Laurent Series with

$\begin{array}{cc}h(z)& =\frac{(z-z_0{)}^m}{z+a}=\frac{(z-z_0{)}^m}{z\underset{=0}{\underbrace{-z_o+z_o}}+a}\\ & =\frac{(z-z_0{)}^m}{(z-z_o)-\underset{=b}{\underbrace{(-z_o-a)}}}=\frac{(z-z_0{)}^m}{(z-z_o)-b}\\ & =\frac{(z-z_0{)}^{m-1}}{1-\frac{b}{z-z_0}}=\frac{(z-z_0{)}^{m-1}}{1-\underset{:=q}{\underbrace{\frac{b}{z-z_o}}}}\\ & =(z-z_0{)}^{m-1}\cdot \sum _{n=0}^{+\mathrm{\infty }}{q}^n=(z-z_0{)}^{m-1}\cdot \sum _{n=0}^{+\mathrm{\infty }}\frac{b^n}{(z-z_o{)}^n}\\ & =(z-z_0{)}^{m-1}\cdot \sum _{n=0}^{+\mathrm{\infty }}{b}^n\cdot (z-z_o{)}^{-n}=(z-z_0{)}^{m-1}\cdot \sum _{n=-\mathrm{\infty }}^0{c}_{-n}\cdot (z-z_o{)}^n\\ & =\sum _{n=-\mathrm{\infty }}^0{c}_{-n}\cdot (z-z_o{)}^{n+m-1}=\sum _{n=-\mathrm{\infty }}^{m-1}{c}_{m-1-n}\cdot (z-z_o{)}^n\end{array}$

The residue for is $n=-1$ erhält man $re{s}_{z_0}(h)=c_{m-1-(-1)}=c_m=b^m=(-z_o-a{)}^m$

See Also

• Theorem on Laurent Series Expansion

Page information

Translation and Version Control

This page was translated based on the following mit Laurentreihen Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

• Source: Kurs:Funktionentheorie/Beispielrechnung mit Laurentreihen - URL:https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Integrationsweg
• Date: 11/20/2024