# 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 $$f(z_0)=\lim_{z\to z_0}(z-z_0)f(z_0)$$.

Created: 2024-05-30 Thu 21:14

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