Complex Analysis/Residuals

Let ${\displaystyle G\subseteq \mathbb {C} }$ be a domain, ${\displaystyle z_{0}\in G}$, and ${\displaystyle f}$ a function that is holomorphic except for isolated singularities ${\displaystyle S\subset G}$, i.e., ${\displaystyle f:G\setminus S\to \mathbb {C} }$ is holomorphic. If ${\displaystyle z_{0}\in S\subset G}$ is an Isolated singularity of ${\displaystyle f}$ with ${\displaystyle D_{r}(z_{0})\cap S={z_{0}}}$, the residue is defined as:

${\displaystyle \mathrm {res} {z_{0}}(f):={\frac {1}{2\pi i}}\int {\partial D_{r}(z_{0})}f(\xi ),d\xi ={\frac {1}{2\pi i}}\int _{|\xi -z_{0}|=r}f(\xi ),d\xi }$.

If ${\displaystyle f}$ is expressed as a Laurent series around an isolated singularity ${\displaystyle z_{0}\in S\subset G}$, the residue can be computed as follows: With ${\displaystyle f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}}$ as the Laurent Expansion of ${\displaystyle f}$ around ${\displaystyle z_{0}}$, it holds:

${\displaystyle \mathrm {res} {z_{0}}(f)={\frac {1}{2\pi i}}\int {\partial D_{r}(z_{0})}f(\xi ),d\xi ={\frac {1}{2\pi i}}\int _{\partial D_{r}(z_{0})}a_{-1}\cdot (\xi -z_{0})^{-1},d\xi ={\frac {1}{2\pi i}}a_{-1}\cdot \underbrace {\int _{\partial D_{r}(z_{0})}(\xi -z_{0})^{-1},d\xi } {=2\pi i}=a{-1}}$.

Here, it is assumed that the closed disk ${\displaystyle {\overline {D_{r}(z_{0})}}}$ contains only the singularity ${\displaystyle z_{0}\in S}$, i.e., ${\displaystyle {\overline {D_{r}(z_{0})}}\cap S={z_{0}}}$.

Thus, the residue ${\displaystyle \mathrm {res} {z_{0}}(f)=a{-1}}$ can be identified as the coefficient of ${\displaystyle (z-z_{0})^{-1}}$ in the Laurent series of ${\displaystyle f}$ around ${\displaystyle z_{0}}$.

The residue (from Latin residuere - to remain) is named so because, during integration along the path ${\displaystyle \gamma (t):=z_{0}+r\cdot e^{it}}$ with ${\displaystyle t\in [0,2\pi ]}$ around ${\displaystyle z_{0}}$, the following holds:

${\displaystyle {\begin{array}{rl}\displaystyle \int _{|w-z_{0}|=r}f(w)\,dw&=\displaystyle \sum _{n=-\infty }^{+\infty }a_{n}\int _{|w-z_{0}|=r}(w-z_{0})^{n}\,dw\\&=2\pi i\cdot a_{-1}\end{array}}}$

The residue is, therefore, what "remains" after integration.

If ${\displaystyle z_{0}\in U}$ is a pole of order ${\displaystyle m}$ of ${\displaystyle f}$, the Laurent Expansion of ${\displaystyle f}$ around ${\displaystyle z_{0}}$ has the form:

${\displaystyle f(z)=\sum _{k=-m}^{\infty }a_{k}(z-z_{0})^{k}}$

with ${\displaystyle a_{-m}\neq 0}$.

By multiplying with ${\displaystyle (z-z_{0})^{m}}$, we obtain:

${\displaystyle g_{m}(z):=(z-z_{0})^{m}\cdot f(z)=\sum _{k=0}^{\infty }a_{k-m}(z-z_{0})^{k}}$

The residue ${\displaystyle a_{-1}}$ is now the coefficient of ${\displaystyle (z-z_{0})^{m-1}}$ in the power series of ${\displaystyle g_{m}(z)}$.

Through ${\displaystyle m-1}$-fold differentiation, the first ${\displaystyle m-1}$ terms in the series, from ${\displaystyle n=0}$ to ${\displaystyle m-2}$, vanish. The residue is then the coefficient of ${\displaystyle (z-z_{0})^{0}}$, yielding:

${\displaystyle g_{m}^{(m-1)}(z)=\sum _{k=m-1}^{\infty }{\frac {k!}{(k-m+1)!}}a_{k-m}(z-z_{0})^{k-m+1}}$.

By shifting the index, we obtain:

${\displaystyle g_{m}^{(m-1)}(z)=\sum _{k=-1}^{\infty }{\frac {m+k!}{(k+1)!}}a_{k}(z-z_{0})^{k+1}}$

Taking the limit ${\displaystyle z\to z_{0}}$, all terms with ${\displaystyle k\geq 0}$ vanish, yielding:

${\displaystyle \lim _{z\to z_{0}}g_{m}^{(m-1)}(z)={\frac {(m-1)!}{0!}}\cdot a_{-1}\cdot (z-z_{0})^{0}=(m-1)!\cdot a_{-1}}$.

Thus, the residue can be computed using the limit ${\displaystyle z\to z_{0}}$:

${\displaystyle \mathrm {res} {z_{0}}(f)=a{-1}={\frac {1}{(m-1)!}}\cdot \lim _{z\to z_{0}}g_{m}^{(m-1)}(z)}$.

• Explain why, during integration of the Laurent series, all terms from the regular part and all terms with index ${\displaystyle n\in \mathbb {Z} }$ with ${\displaystyle n\neq -1}$ contribute

${\displaystyle \int _{\partial D_{r}(z_{0})}a_{n}\cdot (\xi -z_{0})^{n},d\xi =0}$.

• Why is it allowed to interchange the processes of integration and series expansion?

${\displaystyle \sum _{n=-\infty }^{+\infty }\int _{\partial D_{r}(z_{0})}a_{n}(\xi -z_{0})^{n}d\xi =\int _{\partial D_{r}(z_{0})}\sum _{n=-\infty }^{+\infty }a_{n}(\xi -z_{0})^{n}d\xi ={\frac {1}{2\pi i}}\int _{\partial D_{r}(z_{0})}f(\xi ),d\xi =\mathrm {res} _{z_{0}}(f)}$

• Given the function ${\displaystyle f:\mathbb {C} \setminus {i}\to \mathbb {C} }$ with ${\displaystyle z\mapsto f(z)={\frac {e^{z-i}}{(z-i)^{5}}}}$, compute the residue ${\displaystyle \mathrm {res} _{z_{0}}(f)}$ with ${\displaystyle z_{0}:=i}$!.

• Laurent series examples

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

• Source: Kurs:Funktionentheorie/Residuum - URL:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Residuum

• Date: 12/30/2024