Complex Analysis/Example Computation with Laurent Series

The following constants are defined to better illustrate the operations:

• ${\displaystyle b:=-a}$
• ${\displaystyle c:=b-z_{o}=-a-z_{o}}$
• ${\displaystyle c_{n}:={\frac {1}{(b-z_{o})^{n+1}}}={\frac {1}{(-z_{o}-a)^{n+1}}}}$
• ${\displaystyle q:={\frac {z-z_{0}}{c}}}$

Let ${\displaystyle z_{o}\not =-a}$, then:

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

The residue ${\displaystyle res_{z_{0}}(f)=0}$ ,since in the Laurent expansion, the principal part coefficients are all zero (i.e., the principal part vanishes).

• Why is the condition required for the above calculation Laurent Series (or power series)${\displaystyle z_{o}\not =-a}$?
• Compute the Laurent series for ${\displaystyle z_{o}=-a}$ and determine the Residue of the Laurent expansion for ${\displaystyle f(z):={\frac {1}{a+z}}}$ in ${\displaystyle z_{o}=-a}$ at!***

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

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

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

The following constants are defined to better illustrate the operations:

• ${\displaystyle c:=z_{o}-a}$
• ${\displaystyle c_{n}:={\frac {1}{(b+z_{o})^{n+1}}}={\frac {1}{(z_{o}-a)^{n+1}}}}$
• ${\displaystyle q:={\frac {z-z_{o}}{c}}={\frac {z-z_{o}}{z_{o}-a}}}$

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

the residue ${\displaystyle res_{z_{0}}(g)=c_{-1+m}={\frac {1}{(b-z_{o})^{-1+m+1}}}={\frac {1}{(b-z_{o})^{m}}}={\frac {1}{(-z_{o}-a)^{m}}}}$.

The following constants are defined for better clarity:

• ${\displaystyle b:=-z_{0}-a}$
• ${\displaystyle c_{n}:=b^{n}}$
• ${\displaystyle q:={\frac {b}{z-z_{o}}}=-{\frac {z_{0}+a}{z-z_{o}}}}$

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

The residue${\displaystyle res_{z_{0}}(h)=c_{1}=b^{1}=-z_{o}-a}$

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

The residue for is ${\displaystyle n=-1}$ erhält man ${\displaystyle res_{z_{0}}(h)=c_{m-1-(-1)}=c_{m}=b^{m}=(-z_{o}-a)^{m}}$

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