# 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:
* $f:G\to \mathbb {C}$ with
* $G:=\mathbb {C} \setminus \{-a\},a\in \mathbb {C}$
* $f(z):={\frac {1}{a+z}}$
The goal is to develop it into a Laurent series with the expansion point $z_{o}\in \mathbb {C}$..

### 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 $z_{o}\not =-a$, then:
    ${\begin{array}{rl}\displaystyle f(z)&={\frac {1}{a+z}}=\displaystyle -{\frac {1}{b-z}}\ \ (b:=-a)\\&=\displaystyle -{\frac {1}{b\underbrace {-z_{o}+z_{o}} _{=0}-z}}=-{\frac {1}{(b-z_{o})-(z-z_{o})}}\\&=\displaystyle -{\frac {1}{(\underbrace {b-z_{o}} _{=c})-(z-z_{o})}}=-{\frac {1}{c-(z-z_{o})}}\\&=\displaystyle -{\frac {1}{c}}\cdot {\frac {1}{1-{\frac {z-z_{o}}{c}}}}=-{\frac {1}{c}}\cdot {\frac {1}{1-\underbrace {\frac {z-z_{o}}{c}} _{:=q}}}\\&=\displaystyle -{\frac {1}{c}}\cdot {\frac {1}{1-q}}=-{\frac {1}{c}}\cdot \sum _{n=0}^{+\infty }q^{n}=-\sum _{n=0}^{+\infty }{\frac {(z-z_{o})^{n}}{c^{(n+1)}}}\\&=\displaystyle -\sum _{n=0}^{+\infty }{\frac {(z-z_{o})^{n}}{(z_{o}-a)^{n+1}}}=-\sum _{n=0}^{+\infty }\underbrace {\frac {1}{(-z_{o}-a)^{n+1}}} _{c_{n}:=}\cdot (z-z_{o})^{n}\\&=\displaystyle -\sum _{n=0}^{+\infty }c_{n}\cdot (z-z_{o})^{n}\end{array}}$ :
The residue $res_{z_{0}}(f)=0$ ,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)$z_{o}\not =-a$?
* 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:
* $g:G\to \mathbb {C}$ mit
* $G:=\mathbb {C} \setminus \{-a\},a\in \mathbb {C}$
* $g(z):={\frac {1}{(z-z_{o})^{m}\cdot (a+z)}}$
The goal is to develop it into a Laurent series with the expansion point $z_{o}\in \mathbb {C} \setminus \{-a\}$.

### 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

    ${\begin{array}{rl}\displaystyle g(z)&=\displaystyle {\frac {1}{(z-z_{o})^{m}\cdot (a+z)}}={\frac {1}{(z-z_{o})^{m}}}\cdot {\frac {1}{a+z}}\ \ (c:=z_{o}-a)\\&=\displaystyle -{\frac {1}{(z-z_{o})^{m}}}\cdot {\frac {1}{z_{o}-a}}\cdot {\frac {1}{1-{\frac {z-z_{o}}{z_{o}-a}}}}=-{\frac {1}{(z-z_{o})^{m}}}\cdot {\frac {1}{c}}\cdot {\frac {1}{1-q}}\\&=\displaystyle -{\frac {1}{(z-z_{o})^{m}}}\cdot {\frac {1}{c}}\cdot \sum _{n=0}^{+\infty }q^{n}=-\sum _{n=0}^{+\infty }{\frac {(z-z_{o})^{n}}{c^{n+1}}}\\&=\displaystyle -{\frac {1}{(z-z_{o})^{m}}}\cdot \sum _{n=0}^{+\infty }{\frac {(z-z_{o})^{n}}{(z_{o}-a)^{n+1}}}\\&=\displaystyle -{\frac {1}{(z-z_{o})^{m}}}\cdot \sum _{n=0}^{+\infty }\underbrace {\frac {1}{(z_{o}-a)^{n+1}}} _{c_{n}:=}\cdot (z-z_{o})^{n}\\&=\displaystyle -{\frac {1}{(z-z_{o})^{m}}}\cdot \sum _{n=0}^{+\infty }c_{n}\cdot (z-z_{o})^{n}\\&=\displaystyle -\sum _{n=-m}^{+\infty }c_{n+m}\cdot (z-z_{o})^{n}\\\end{array}}$
the residue $res_{z_{0}}(g)=c_{-1+m}={\frac {1}{(b-z_{o})^{-1+m+1}}}={\frac {1}{(b-z_{o})^{m}}}={\frac {1}{(-z_{o}-a)^{m}}}$.

## Laurent Series with Infinite Principal Part Terms

A simple rational function of the following form is given:
* $h:G\to \mathbb {C}$ with
* $G:=\mathbb {C} \setminus \{-a\},a\in \mathbb {C} \setminus \{-a\}$
* $h(z):={\frac {(z-z_{0})^{m}}{z+a}}$
The goal is to develop it into a Laurent series with the expansion point $z_{o}\in \mathbb {C} \setminus \{-a\}$.

### 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

    ${\begin{array}{rl}\displaystyle h(z)&=\displaystyle {\frac {z-z_{0}}{z+a}}={\frac {z-z_{0}}{z\underbrace {-z_{o}+z_{o}} _{=0}+a}}\\&=\displaystyle {\frac {z-z_{0}}{(z-z_{o})-\underbrace {(-z_{o}-a)} _{=b}}}={\frac {z-z_{0}}{(z-z_{o})-b}}\\&=\displaystyle {\frac {1}{1-{\frac {b}{z-z_{0}}}}}={\frac {1}{1-\underbrace {\frac {b}{z-z_{o}}} _{:=q}}}\\&=\displaystyle \sum _{n=0}^{+\infty }q^{n}=\sum _{n=0}^{+\infty }{\frac {b^{n}}{(z-z_{o})^{n}}}\\&=\displaystyle \sum _{n=0}^{+\infty }b^{n}\cdot (z-z_{o})^{-n}=\sum _{n=-\infty }^{0}c_{-n}\cdot (z-z_{o})^{n}\end{array}}$
The residue$res_{z_{0}}(h)=c_{1}=b^{1}=-z_{o}-a$

### Transformation into a Laurent Series with

    ${\begin{array}{rl}\displaystyle h(z)&=\displaystyle {\frac {(z-z_{0})^{m}}{z+a}}={\frac {(z-z_{0})^{m}}{z\underbrace {-z_{o}+z_{o}} _{=0}+a}}\\&=\displaystyle {\frac {(z-z_{0})^{m}}{(z-z_{o})-\underbrace {(-z_{o}-a)} _{=b}}}={\frac {(z-z_{0})^{m}}{(z-z_{o})-b}}\\&=\displaystyle {\frac {(z-z_{0})^{m-1}}{1-{\frac {b}{z-z_{0}}}}}={\frac {(z-z_{0})^{m-1}}{1-\underbrace {\frac {b}{z-z_{o}}} _{:=q}}}\\&=(z-z_{0})^{m-1}\cdot \displaystyle \sum _{n=0}^{+\infty }q^{n}=(z-z_{0})^{m-1}\cdot \sum _{n=0}^{+\infty }{\frac {b^{n}}{(z-z_{o})^{n}}}\\&=(z-z_{0})^{m-1}\cdot \displaystyle \sum _{n=0}^{+\infty }b^{n}\cdot (z-z_{o})^{-n}=(z-z_{0})^{m-1}\cdot \sum _{n=-\infty }^{0}c_{-n}\cdot (z-z_{o})^{n}\\&=\displaystyle \sum _{n=-\infty }^{0}c_{-n}\cdot (z-z_{o})^{n+m-1}=\displaystyle \sum _{n=-\infty }^{m-1}c_{m-1-n}\cdot (z-z_{o})^{n}\end{array}}$
The residue for is $n=-1$ erhält man $res_{z_{0}}(h)=c_{m-1-(-1)}=c_{m}=b^{m}=(-z_{o}-a)^{m}$

## 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