30 Jul 2025

Taylor and Laurent Expansions, Residues, and the Residue Theorem

Series from Cauchy's formula, annular Laurent expansions, residue calculation, and contour sums over enclosed poles.

bsc semester-ii mathematical-physics laurent-series residue-theorem

Taylor expansion

Let $f$ be analytic inside a circle containing $z$ and centered at $z_0$. Cauchy’s formula is

\[f(z)=\frac{1}{2\pi i} \oint_C\frac{f(\zeta)}{\zeta-z}d\zeta.\]

Write

\[\frac1{\zeta-z} =\frac1{\zeta-z_0} \frac1{1-\dfrac{z-z_0}{\zeta-z_0}}.\]

When $\lvert z-z_0\rvert<\lvert\zeta-z_0\rvert$, the geometric series gives

\[\frac1{\zeta-z} =\sum_{n=0}^{\infty} \frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}}.\]

Substitution into Cauchy’s formula yields

\[f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n,\]

where

\[a_n=\frac{1}{2\pi i} \oint_C\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta =\frac{f^{(n)}(z_0)}{n!}.\]

Thus

\[\boxed{ f(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n}.\]

Laurent expansion

If $f$ is analytic only in an annulus

\[r_1<\lvert z-z_0\rvert<r_2,\]

negative as well as nonnegative powers may occur:

\[\boxed{ f(z)=\sum_{n=-\infty}^{\infty} a_n(z-z_0)^n}.\]

The coefficients have the single contour formula

\[\boxed{ a_n=\frac{1}{2\pi i} \oint_C\frac{f(\zeta)} {(\zeta-z_0)^{n+1}}d\zeta},\]

where $C$ is any positively oriented circle inside the annulus.

For example,

\[\frac1{z(1-z)} =\frac1z(1+z+z^2+\cdots) =\frac1z+1+z+z^2+\cdots\]

for $0<\lvert z\rvert<1$. For $\lvert z\rvert>1$,

\[\frac1{z(1-z)} =-\frac1{z^2}\frac1{1-1/z} =-\frac1{z^2}-\frac1{z^3}-\cdots.\]

The same function has different Laurent expansions in different annuli.

A Laurent series belongs to an annulus bounded by singularities. The residue at \(z_0\) is the coefficient \(a_{-1}\), isolated by one positive circuit around the center.

Residues

The residue of $f$ at an isolated singularity $z_0$ is

\[\boxed{\operatorname{Res}(f,z_0)=a_{-1}}.\]

For a simple pole,

\[\boxed{ \operatorname{Res}(f,z_0) =\lim_{z\to z_0}(z-z_0)f(z)}.\]

For a pole of order $m$,

\[\boxed{ \operatorname{Res}(f,z_0) =\frac1{(m-1)!} \lim_{z\to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[(z-z_0)^mf(z)\right]}.\]

Residue theorem

Let $f$ be analytic inside and on a positively oriented contour $C$ except for isolated singularities $z_1,\ldots,z_N$. Remove small positively oriented circles around them. Cauchy-Goursat on the remaining region gives the outer integral equal to the sum of the small-circle integrals:

\[\oint_Cf(z)dz =\sum_{k=1}^N\oint_{C_k}f(z)dz.\]

In each Laurent series, every term integrates to zero except $a_{-1}(z-z_k)^{-1}$:

\[\oint_{C_k}(z-z_k)^n dz = \begin{cases} 2\pi i,&n=-1,\\ 0,&n\ne-1. \end{cases}\]

Therefore

\[\boxed{ \oint_C f(z)dz =2\pi i\sum_{k=1}^N\operatorname{Res}(f,z_k)}.\]

The Laurent coefficients and sample residues are checked in the Unit III Maxima worksheet.