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?


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\(_n\) \(g(z_0) = \frac{f^{(n)}(z_0)}{n!}\).


  • \(\gamma\) 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 \(f(z_0)=\lim_{z\to z_0}(z-z_0)f(z_0)\).

Author: Christian Cunningham

Created: 2024-05-30 Thu 21:14