# Residues

(Lecture 23 of Mathematical Methods II.)

Residue theory is the culmination of complex integration, bringing together Cauchy's integral formula and Laurent Series. Integrating both sides of the Laurent expansion around $z_0$:

$$\tag{1} \oint f(z)\,dz=\sum_{n=-\infty}^\infty c_n\oint (z-z_0)^n\,dz\,,$$

and remembering that $\oint(z-z_0)^n\,dz=2i\pi\delta_{n,-1}$, we find:

$$\tag{2} \oint f(z)\,dz=2i\pi c_{-1}$$

The coefficient of the negative power of order one in the Laurent expansion is called the Residue. We write: $\mathrm{Res}_{z=z_0}f(z)=c_{-1}$. To calculate a residue at a pole, we need not produce a whole Laurent series, but, more economically, we can derive formulas to compute residues:

Simple poles:

\begin{align} \tag{3} \mathrm{Res}_{z=z_0}f(z)&=\lim_{z\rightarrow z_0}(z-z_0)f(z)\,,\\ \mathrm{Res}_{z=z_0}\frac{p(z)}{q(z)}&=\frac{p(z_0)}{q'(z_0)}\,, \end{align}

where, in the latter case, $f(z_0)\neq0$ and $z_0$ is a simple zero of $q$.

The generalization to higher poles provides the formula:

$$\tag{4} \mathrm{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\lim_{z\rightarrow z_0}\left\{\frac{d^{m-1}}{dz^{m-1}}\left[(z-z_0)^m f(z)\right]\right\}\,.$$

In particular, for a pole of order 2:

$$\mathrm{Res}_{z=z_0}f(z)=\lim_{z\rightarrow z_0}\{[(z-z_0)^2f(z)']\tag{5}\,.$$

The so-called Residue Theorem extends this technique to the case of several singularities: if $f$ is holomorphic inside a simple closed path $\mathcal{C}$ and on $\mathcal{C}$, except for finitely many points $z_1$, \dots, $z_n$, inside $\mathcal{C}$, then the integral of $\mathcal{C}$ in the trigonometric sense is $2i\pi$ times the sum of the residues at the poles:

$$\tag{6} \oint_\mathcal{C}f(z)\,dz=2i\pi\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z)\,,$$

The method of Residue is useful to compute integrals not only in the complex planes but also of the real variable. For instance, rational functions of $\cos(\theta)$ and $\sin(\theta)$ in polar coordinates are easily obtained through the substitution:

\begin{align} \tag{7} \cos(\theta)&=\frac{1}{2}\left(z+\frac{1}{z}\right)\,,\\ \sin(\theta)&=\frac{1}{2i}\left(z-\frac{1}{z}\right)\,, \end{align}

that turns them into rational functions of $z$ on a complex circle; indeed, $d\theta=dz/(iz)$ and:

$$\tag{8} \int_0^{2\pi} F(\cos(\theta),\sin(\theta))\,d\theta=\oint_\mathcal{C}\frac{f(z)}{iz}\,dz\,,$$

that can thus be integrated by the method of Residues. We will see in next lectures how other types of integrals can be similarly easily calculated thanks to residues.