Cauchy Residues

Recall that $0 = \oint_{γ} d z f (z) = 0$ if $f (z)$ is holomorphic in and on $γ$ .

What if it isn’t?

Residues

Res $f (z_{0}) = \frac{1}{2 π i} \oint_{γ} d z f (z)$ , Residue of $f$ at $z_{0}$ .

Residue for Arbitrary Pole - Guess

Recall $f^{(n)} (z) = \frac{n!}{2 π i} \oint d z^{'} \frac{f (z^{'})}{(z^{'} - z)^{n + 1}}$

Let $g (z) = \frac{f (z)}{(z - z_{0})^{n + 1}}$ such that, $g (z)$ has a Lauraunt series with a non-zero principle part and $f (z)$ only has a taylor series.

So, the residue of an $n$-th order pole is Res $_{n}$ $g (z_{0}) = \frac{f^{(n)} (z_{0})}{n!}$ .

Assumptions

$γ$ is CCW
$γ$ wraps around only once - winding number of 1, i.e. simple
$z_{0}$ is the only singularity inside $γ$ .

Residue Theorem

Suppose $f (z)$ is meromorphic in a region bounded by a contour with singular points ${z_{1}, \dots, z_{n}}$ inside $γ$ . Then, $\oint_{γ} d z^{'} f (z^{'}) = 2 π i \sum_{z_{i}} Res f (z_{i})$ .

Alternative: Suppose $f (z)$ is meromorphic inside and on a simple closed contour $γ$ with signular points ${z_{1}, \dots, z_{n}}$ inside $γ$ . Then, $\oint_{γ} d z^{'} f (z^{'}) = 2 π i \sum_{z_{i}} Res f (z_{i})$ .

Thus, if we can compute Residues more efficiently than an integral, then we can compute an integral without doing it directly.

First Coefficient of Laurent Series as Residue

$b_{1} = \frac{1}{2 π i} \oint d z^{'} f (z^{'}) = Res f (z_{0})$ . I.e. the coefficient of the $z - z_{0}$ term in the principle part.

In general

Suppose $f (z)$ has a pole of order $n$ . $f (z) = \sum_{m} a_{m} (z - z_{0})^{m} + \sum_{m = 1}^{n} \frac{b_{m}}{(z - z_{0})^{m}}$ . Then, let $g (z)$ be such that $f (z) = \frac{g (z)}{(z - z_{0})^{n}}$ . I.e. $g (z) = \sum_{m} a_{m} (z - z_{0})^{m + n} + \sum_{m = 1}^{n} \frac{b_{m}}{(z - z_{0})^{m - n}}$ . Thus, $g (z)$ is holomorphic around $z_{0}$ and is nonzero at $z_{0}$ . Then, Res $f (z_{0}) = \frac{1}{2 π i} \oint d z^{'} f (z^{'}) = \frac{1}{2 π i} \oint d z^{'} \frac{g (z^{'})}{(z^{'} - z_{0})^{n}} = \frac{1}{(n - 1)!} \frac{d^{n - 1} g}{d z^{n - 1}} (z_{0})$ .

Residue General Formula for n-Pole

So, Res $f (z_{0}) = lim_{z \to z_{0}} \frac{1}{(n - 1)!} \frac{d^{n - 1}}{d z^{n - 1}} (z - z_{0})^{n} f (z)$ . For a simple pole, $n = 1$ , hence Res $f (z_{0}) = lim_{z \to z_{0}} (z - z_{0}) f (z_{0})$ .