Laurant Expansion

Laurent expansion, represents a complex function as a power series that includes terms of negative degree. Unlike Taylor series, which only considers positive powers, the Laurent series accounts for both positive and negative powers.

$$ f(z)=\sum_{n=-\infty}^{\infty} a_n(z-c)^n $$

The part of the series that contains negative powers is called the principal part.

$$ \oint_C f(z)dz=\sum_{n=-\infty}^\infty a_n\oint_C (z-c)^ndz $$ When $c$ is an isolated singularity, the following holds. $$ \oint_C (z-c)^n dz= \begin{cases} 2\pi i & n=-1\\ 0 & n\neq -1 \end{cases} $$

$a_{-1}$ is called residue.

$$ \begin{align*} a_{-1}=\operatorname{Res}_{z=c} f(z)&={1\over 2\pi i}\oint_C f(z)dz\\ \oint_C f(z)dz &= 2\pi i\cdot \operatorname{Res}_{z=c} \end{align*} $$

When the isolated singularity is a pole of order $n$,

$$ (z-c)^n f(z)=\sum_{k=0}^\infty a_{k-n}(z-c)^k $$ and

$$ \operatorname{Res}_{z=c} f(z)={1\over (n-1)!}\lim_{z\to c}{d^{n-1}\over dz^{n-1}}[(z-c)^n f(z)] $$

Let us calculate the integral over a circle centered at $i$ with radius 1.

$$ f(z)={z^2\over (z-i)} $$

This function has an isolated singularity at $z=i$ and the singularity is a pole of order 1.

$$ \begin{align*} \operatorname{Res}_{z=i} f(z)&=\lim_{z\to i}(z-i)\cdot {z^2\over (z-i)}=-1\\ \oint_C f(z)dz &= -2\pi i \end{align*} $$

from sympy import residue, I
from sympy.abc import z

residue(z**2/(z-I),z,I)

$\displaystyle -1$

Let us calculate the integral over a circle centered at origin with radius 10.

$$ f(z)={1\over z(z-1)^2} $$

This function has an isolated singularity at $z=0$ and the singularity is a pole of order 1 and has another isolated singularity at $z=1$ and the singularity is a pole of order 2. $$ \begin{align*} \operatorname{Res}_{z=0} f(z)&=\lim_{z\to 0} z\cdot {1\over z(z-1)^2}\\ &=1\\ \operatorname{Res}_{z=1} f(z)&=\lim_{z\to 1} {d\over dz}(z-1)^2\cdot {1\over z(z-1)^2}\\ &=\lim_{z\to 1}(-{1\over z^2})\\ &=-1\\ \oint {1\over z(z-1)^2} dz &= 2\pi i(1-1)\\ &=0 \end{align*} $$

residue(1/(z*(z-1)**2),z,0)

$\displaystyle 1$

residue(1/(z*(z-1)**2),z,1)

$\displaystyle -1$

Let us calculate the integral over a half circle centered at origin with radius 10 in the positive plane.

$$ f(z)={z-2\over (z^2+1)^2} $$

This function has an isolated singularity at $z=i$ and the singularity is a pole of order 2.

$$ \begin{align*} \operatorname{Res}_{z=i} f(z)&=(z-i)^2\cdot {z-2\over (z^2+1)^2}\\ &=\lim_{z\to i}{d\over dz}{z-2\over (z+i)^2}\\ &=\lim_{z\to i}{(z+i)^2-2(z+i)(z-2)\over (z+i)^4}\\ &=-{1\over 2i}\\ \oint_C f(z)dz &= 2\pi i\cdot -{1\over 2i}\\ &=-\pi \end{align*} $$

residue((z-2)/((z**2+1)**2),z,I)

$\displaystyle \frac{i}{2}$

import sympy 
sympy.diff((z-2)/(z+I)**2,z)

$\displaystyle - \frac{2 \left(z - 2\right)}{\left(z + i\right)^{3}} + \frac{1}{\left(z + i\right)^{2}}$