Residue

Our next topic in the course is based on the idea of using isolated singularities and Laurent expansions to evaluate contour integrals, which goes by the name “Residue Theory.” This is a very widely applicable toolkit that gets used extensively in parts of physics and electrical engineering.

The Residue Theorem

Residue

If and , then the residue of at is for any . The residue is independent of the choice of by homotopy invariance. Note that the residue is just the coefficient of in the Laurent expansion of centered at .

Definition

If is a domain, and , then means there exists an open set containing such that has a holomorphic extension to .

Residue Theorem

Let be a bounded Jordan domain, let , and let . Then

Residue Theorem Generalises Cauchy Integral Formula

This is a generalization of the Cauchy integral formula, which says that the integral on the left-hand side is zero if there are no singularities inside the domain.

Proof Choose some , such that for all , and let Clearly is a Jordan domain. Then by Cauchy integral formula, we have where the minus sign arises from different orientations. Since by definition of residue, we have for all , it follows that $$\int_{\partial D}f(z)\dd{z}=2\pi i\sum_{j=1}^n\operatorname{Res}_{a_j}f$$$\square$

e.g. Let be a bounded Jordan domain and let . Then at each , we have the Laurent series: Take the derivative, we have the coefficient of disappear. That is, the residue of at is zero. Therefore, by the residue theorem, we have which aligns with the fact that, is a primitive of , so the closed path integral of is zero.

Residue Computation

Proposition

Suppose and are holomorphic on a neighborhood of , with and of order . Then

Proof Since and has a simple zero at , we can write the Taylor expansions: with . is holomorphic on , so we can writeThen we have Note that the residue of at is the coefficient of , that is . Since and , the result follows.

e.g.

• Let and . has a simple zero at , apply the proposition, we have Furthermore, by the residue theorem, we have

Contour Integration

We will now see some examples to illustrate how to apply these techniques to compute certain kinds of real integrals in surprising and powerful ways:

Proposition

Let be the upper half-plane. Suppose we have a function and with satisfying , parameterising the upper semicircle of radius . Then

Proof Let be the concatenation of and the line segment , i.e. is the boundary of the upper half-disk of radius . Then by the residue theorem, we have $$\begin{aligned}&\int_{-\infty}^{\infty}f(x) \dd x\ =&\lim_{R\to\infty}\int_{[-R,R]}f(z)\dd{z}\=&\lim_{R\to \infty}\left(\int_{[-R,R]}f(z)\dd{z}+\int_{\gamma_{R}}f(z)\dd{z}\right)\=&\lim_{R\to \infty}\int_{\rho_{R}}f(z)\dd{z}\=&2\pi i\sum_{j=1}^{n}\res_{a_j}f\end{aligned}$$$\square$

Corollary

Suppose where and are polynomials with for all and , all of the roots of in the upper half-plane have multiplicity . Then where are the roots of in the upper half-plane.

Proof Since , we have for some real constant . Therefore, by ML estimation, Thus we can apply the proposition, and get Notice that the residue is given by the proposition, and the result follows.

e.g.

• since has only one root in the upper half-plane.
• Consider . The roots of in the upper half-plane are given by . Hence Therefore, the integral is