Cauchy Residues

Recall that $0=\oint_\gamma dz f(z)=0$ if $f(z)$ is holomorphic in and on $\gamma$.

What if it isn’t?

Residues

Res $f(z_0)=\frac{1}{2\pi i}\oint_\gamma dz f(z)$, Residue of $f$ at $z_0$.

Residue for Arbitrary Pole - Guess

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

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

$\gamma$ is CCW
$\gamma$ wraps around only once - winding number of 1, i.e. simple
$z_0$ is the only singularity inside $\gamma$.

Residue Theorem

Suppose $f(z)$ is meromorphic in a region bounded by a contour with singular points $\{z_1,\cdots,z_n\}$ inside $\gamma$. Then, $\oint_\gamma dz' f(z') = 2\pi i\sum_{z_i}\text{Res } f(z_i)$.

Alternative: Suppose $f(z)$ is meromorphic inside and on a simple closed contour $\gamma$ with signular points $\{z_1,\cdots,z_n\}$ inside $\gamma$. Then, $\oint_\gamma dz' f(z') = 2\pi i\sum_{z_i}\text{Res } f(z_i)$.

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

$b_1=\frac{1}{2\pi i}\oint dz'f(z') = \text{Res }f(z_0)$. 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 $f(z_0) = \frac{1}{2\pi i}\oint dz' f(z') = \frac{1}{2\pi i}\oint dz' \frac{g(z')}{(z'-z_0)^n} = \frac{1}{(n-1)!}\frac{d^{n-1}g}{dz^{n-1}}(z_0)$.

Residue General Formula for n-Pole

So, Res $f(z_0) = \lim_{z\to z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)$. For a simple pole, $n=1$, hence Res $f(z_0)=\lim_{z\to z_0}(z-z_0)f(z_0)$.