Residue

Definition

Let $G\subseteq \mathbb {C}$ be a domain, $z_{0}\in G$ , and let $f$ be holomorphic except for isolated singularities $S\subset G$ , i.e., $f\colon G\setminus S\to \mathbb {C}$ 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 {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

.

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 $f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}$ as the Laurent Series expansion of $f$ around $z_{0}$ , it holds that:

\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}

.

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' $\mathrm {res} _{z_{0}}(f)=a_{-1}$ 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 $\mathrm {res} {z_{0}}(f)=a{-1}$ 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}{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}}

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}^{\infty }a_{k}(z-z_{0})^{k}

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

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}^{\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}^{\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}^{\infty }{\frac {(m+k)!}{(k+1)!}}a_{k}(z-z_{0})^{k+1}.

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

\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 $z\to z_{0}$ using:

\mathrm {res} {z_{0}}(f)=a{-1}={\frac {1}{(m-1)!}}\cdot \lim _{z\to z_{0}}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 $n\neq -1$ , yield integrals that evaluate to zero:

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

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

\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 $f:\mathbb {C} \setminus {i}\to \mathbb {C}$ with $z\mapsto f(z)={\frac {e^{z-i}}{(z-i)^{5}}}$ , calculate the residue $\mathrm {res} _{z_{0}}(f)$ at $z_{0}:=i$ .

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Date: 12/26/2024