Residue

Definition

Let $G\subseteq ℂ$ be a domain, $z_0\in G$, and let $f$ be holomorphic except for isolated singularities $S\subset G$, i.e., $f:G\setminus S\to ℂ$ is holomorphic. If $z_0\in S\subset G$ is an Isolated singularity of $f$ with $D_r(z_0)\cap S=z_0$, the residue is defined as:

$\mathrm{r}\mathrm{e}\mathrm{s}{z}_0(f):=\frac{1}{2\pi i}\int \partial D_r(z_0)f(\xi ),d\xi =\frac{1}{2\pi i}\underset{|\xi -z_0|=r}{\int }f(\xi ),d\xi$.

Relation Between Residue and Laurent Series

If $f$ is represented around an isolated singularity $z_0\in S\subset G$ as a Laurent series, the residue can be computed as follows. With $f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$ as the Laurent Series expansion of $f$ around $z_0$, it holds that:

$\mathrm{r}\mathrm{e}\mathrm{s}{z}_0(f)=\frac{1}{2\pi i}\int \partial D_r(z_0)f(\xi ),d\xi =\frac{1}{2\pi i}\underset{\partial D_r(z_0)}{\int }{a}_{-1}\cdot (\xi -z_0{)}^{-1},d\xi =\frac{1}{2\pi i}{a}_{-1}\cdot \underbrace{\underset{\partial D_r(z_0)}{\int }(\xi -z_0{)}^{-1},d\xi }=2\pi i=a-1$.

It must be taken into account that the closed disk $\overline{D_r(z_0)}$ contains only the singularity $z_0\in S$, i.e$\overline{D_r(z_0)}\cap s=\{z_0\}$.

Thus, one can read off the 'residue' ${\mathrm{r}\mathrm{e}\mathrm{s}}_{z_0}(f)=a_{-1}$ from the Laurent expansion of around at the $f$ um $z_0$ an -1-ten coefficient of .

Considerations

The closed disk $\overline{D_r(z_0)}$ must contain only the singularity $z_0\in S$, meaning$\overline{D_r(z_0)}\cap S=z_0$. Thus, the residue $\mathrm{r}\mathrm{e}\mathrm{s}{z}_0(f)=a-1$ can be read off directly as the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series expansion of $f$ around $z_0$.

Etymology

The term "residue" (from Latin residuere – to remain) is used because in integration along the path $\gamma (t):=z_0+r\cdot e^{it}$ with $t\in [0,2\pi ]$ around the circle centered at $z_0$, the following holds:

$\begin{array}{cc}\underset{|w-z_0|=r}{\int }f(w)dw& =\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{a}_n\underset{|w-z_0|=r}{\int }(w-z_0{)}^{n}dw\\ & =2\pi i\cdot a_{-1}.\end{array}$

Thus, the residue is what remains after integrating.

Computation for Poles

If $z_0\in U$ is a pole of order $m$ of $f$, the Laurent Series expansion of $f$ around $z_0$ has the form:

$f(z)=\sum _{k=-m}^{\mathrm{\infty }}{a}_k(z-z_0{)}^k$

with $a_{-m}\ne 0$.

Proof 1: Removing Principal Part by Multiplication

By multiplying with $(z-z_0{)}^m$, we get:

$g_m(z):=(z-z_0{)}^m\cdot f(z)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k-m}(z-z_0{)}^k.$

The residue $a_{-1}$ is then the coefficient of $(z-z_0{)}^{m-1}$ in the power series of $g_m(z)$.

Proof 2: Using (m-1)-fold Differentiation

By differentiating $m-1$ times, the first $m-1$ terms from $n=0$ to $n=m-2$ vanish. The residue is then found as the coefficient of $(z-z_0{)}^0$ in:

$g_m^{(m-1)}(z)=\sum _{k=m-1}^{\mathrm{\infty }}\frac{k!}{(k-m+1)!}{a}_{k-m}(z-z_0{)}^{k-m+1}.$

Proof 3: Limit Process to Compute Coefficient of $(z-z_0{)}^0$

By shifting the index:

$g_m^{(m-1)}(z)=\sum _{k=-1}^{\mathrm{\infty }}\frac{(m+k)!}{(k+1)!}{a}_k(z-z_0{)}^{k+1}.$

Taking the limit $z\to z_0$, all terms with $k\ge 0$ vanish, leaving:

$\underset{z\to z_0}{\lim }{g}_m^{(m-1)}(z)=\frac{(m-1)!}{0!}\cdot a_{-1}\cdot (z-z_0{)}^0=(m-1)!\cdot a_{-1}.$

Thus, the residue can be computed $z\to z_0$ using:

$\mathrm{r}\mathrm{e}\mathrm{s}{z}_0(f)=a-1=\frac{1}{(m-1)!}\cdot \underset{z\to z_0}{\lim }{g}_m^{(m-1)}(z).$

Exercises for Students

• Explain why, in the Laurent series expansion, all terms from the principal and outer parts, i.e., $n\in \mathbb{Z}$ with $n\ne -1$, yield integrals that evaluate to zero:

$\underset{\partial D_r(z_0)}{\int }{a}_n\cdot (\xi -z_0{)}^n,d\xi =0.$

• Why can the order of integration and series expansion be interchanged?

$\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}\underset{\partial D_r(z_0)}{\int }{a}_n(\xi -z_0{)}^{n}d\xi =\underset{\partial D_r(z_0)}{\int }\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{a}_n(\xi -z_0{)}^{n}d\xi =\frac{1}{2\pi i}\underset{\partial D_r(z_0)}{\int }f(\xi ),d\xi ={\mathrm{r}\mathrm{e}\mathrm{s}}_{z_0}(f).$

• Given the function $f:ℂ\setminus i\to ℂ$ with $z\mapsto f(z)=\frac{e^{z-i}}{(z-i{)}^5}$, calculate the residue ${\mathrm{r}\mathrm{e}\mathrm{s}}_{z_0}(f)$ at $z_0:=i$.

See Also

• Residue
• Examples of Laurent Series

Page Information

Translation and Version Control

This page was translated based on the following [https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Residuum 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: 12/26/2024

de:Kurs:Funktionentheorie/Residuum