16 - Residue

Residue

Let $z_0$ be an isolated singularity of $f$ then

Residue of $f$ at $z_0$ is defined as the coefficient of $c_{-1}$ of the Lareunt Expansion denoted by

$${\mathrm{Res}}_{z_0}f\text{ or }\mathrm{Res}(f,z_0)$$

Reason for such Definition

Consider function $f$ given by series $\sum c_n(z-z_0{)}^n$
which converges uniformly in an annulus containing ${\gamma }_s$ then integrate term by term

For $n\ne 1$ then $(z-z_0{)}^n$ has a well defined primitive $\frac{(z-z_0{)}^{n+1}}{n+1}$ so integrate to $0$ on ${\gamma }_s$

However for $n=-1$ then the integral of $1/(z-z_0)$ along ${\gamma }_s$ is $2\pi i$ so

$$\frac{1}{2\pi i}{\int }_{{\gamma }_s}f(z)dz=\frac{1}{2\pi i}{\int }_{{\gamma }_s}\sum _{n=-\infty }^{\infty }{c}_n(z-z_0{)}^n=\frac{1}{2\pi i}{\int }_{{\gamma }_s}\frac{c_{-1}}{z-z_0}dz=c_{-1}$$

Residue of the Logarithmic Derivative lemma

Suppose $f:U\to \mathbb{C}$ is meromorphic and has zero of order $k$ or pole of order $k$ at $z_0\in U$ then

$$\frac{f^{\prime }(z)}{f(z)}\text{ has a simple pole at }{z}_0\text{ with residue }k\text{ or }-k\text{ respectively}$$

Proof

If $f(z)$ has a zero of order $k$ then

$$f(z)=(z-z_0{)}^{k}g(z)$$

where $g(z)$ is holomorphic near $z_0$ and $g(z_0)\ne 0$ then

$$\frac{f^{\prime }(z)}{f(z)}=\frac{k}{z-z_0}+\frac{g^{\prime }(z)}{g(z)}$$

since $g(z)\ne 0$ near $z_0$ then $\frac{g^{\prime }(z)}{g(z)}$ is holomorphic near $z_0$ so result follows

Case is similar where $f$ has pole at $z_0$

Residue of a Simple Positively Oriented Contour

Let $\gamma$ be a simple positively oriented contour

Let $f$ be holomorphic on and inside of $\gamma$ except at a finite set $S$ of isolated singularities
where none of the singularities lie on ${\gamma }^{\ast }$ itself then

$${\int }_{\gamma }f(z)dz=2\pi i\sum _{z_0\in S}{\mathrm{Res}}_{z_0}(f)$$

Residue of Poles

Suppose $f$ has a pole of order $m$ at $z_0$ then

$${\mathrm{Res}}_{z_0}(f)=\underset{z\to z_0}{\lim }\frac{1}{(m-1)!}\frac{d^{m-1}}{d{z}^{m-1}}((z-z_0{)}^{m}f(z))$$

Residue of Poles for Ratios $f=\frac{g}{h}$ is a ratio of two holomorphic functions defined on domain $U\subseteq \mathbb{C}$ where $h$ is non-constant then

Suppose

$f$ is meromorphic with poles at zeros of $h$

If $h$ has a simple zero at $z_0$ and $g$ is non-vanishing then $f$ has a simple pole at $z_0$
Since the zero of $h$ is simple at $z_0$ then $h^{\prime }(z_0)\ne 0$ hence

$${\mathrm{Res}}_{z_0}(f)=\underset{z\to z_0}{\lim }\frac{g(z)(z-z_0)}{h(z)}=\underset{z\to z_0}{\lim }g(z)\underset{z\to z_0}{\lim }\frac{z-z_0}{h(z)-h(z_0)}=\frac{g(z_0)}{h^{\prime }(z_0)}$$

Proof

Since $f$ has a pole of order $m$ at $z_0$ then

$$f(z)=\sum _{n\ge -m}{c}_n(z-z_0{)}^n$$

where $z$ is sufficiently close to $z_0$ hence

$$(z-z_0{)}^{m}f(z)=c_{-m}+c_{-m+1}(z-z_0)+\cdots +c_{-1}(z-z_0{)}^{m-1}+\cdots$$

where the result follows from the formula derivatives of a power series

Boundary of Half-Disk Contour Integral

Suppose we want to find integral of

$${\int }_0^{\infty }f(x)dx$$

Define contour ${\Gamma }_R$ as the concatenation of paths ${\gamma }_1:[-R,R]\to \mathbb{C}$ and ${\gamma }_2:[0,1]\to \mathbb{C}$ where

$${\gamma }_1(t)=-R+t\text{ and }{\gamma }_2(t)={\mathrm{Re}}^{i\pi t}$$

such that ${\Gamma }_R={\gamma }_2\star {\gamma }_1$ traces boundary of half-disk

In many cases it can be shown that

$${\int }_{{\gamma }_2}f(z)dz\to 0\text{ as }R\to \infty$$

So by calculating the residues inside contour ${\Gamma }_R$ then the
Integral of $f$ on $(-\infty ,\infty )$ can be deduced

Residue from Small Circle Integrals lemma

Let $f:U\to \mathbb{C}$ be a meromorphic function with a simple pole at $a\in U$
Let ${\gamma }_{ϵ}:[\alpha ,\beta ]\to \mathbb{C}$ be path defined by

$${\gamma }_{ϵ}(t)=a+ϵe^{it}$$

then

$$\underset{ϵ\to 0}{\lim }{\int }_{{\gamma }_{ϵ}}f(z)dz={\mathrm{Res}}_a(f)(\beta -\alpha )i$$

Proof

Since $f$ has a simple pole at $a$ then

$$f(z)=\frac{c}{z-a}+g(z)$$

where $g(z)$ is holomorphic near $z$ and $c={\mathrm{Res}}_a(f)$

Note that $\frac{c}{z-a}$ is the principal part of $f$ at $a$

As $g$ is holomorphic at $a$ then it is continuous at $a$ and hence bounded
Let $M,r>0$ such that

$$|g(z)|\le M\text{ for all }z\in B(a,r)$$

Then if $0<ϵ<r$ then

$$\left|{\int }_{{\gamma }_{ϵ}}g(z)dz\right|\le \ell ({\gamma }_e)M=(\beta -\alpha )ϵM$$

which tends to $0$ as $ϵ\to 0$

However also

$${\int }_{{\gamma }_{ϵ}}\frac{c}{z-a}dz={\int }_{\alpha }^{\beta }\frac{c}{ϵe^{it}}iϵe^{it}dx={\int }_{\alpha }^{\beta }(ic)dt=ic(\beta -\alpha )$$

Since

$${\int }_{{\gamma }_{ϵ}}f(z)dz={\int }_{{\gamma }_{ϵ}}\frac{c}{z-a}dz+{\int }_{{\gamma }_{ϵ}}g(z)dz$$

then result follows

Keyhole Contours

Constructed from two circular paths of radius $ϵ$ and $R$
where $R$ becomes arbitrarily large and $ϵ$ arbitrarily small

Join the two circles by line segments with narrow neck in between
Explicitly if $0<ϵ<R$ then pick $\delta >0$ with

$$\begin{array}{rl}{\eta }_{+}(t)& =t+i\delta \\ {\eta }_{-}(t)& =(R-t)-i\delta \end{array}$$

where $t$ runs over closed interval with endpoints such that
Endpoints of ${\ell }_{\pm }$ lie on circles of radius $ϵ$ and radius $R$ about origin

Letting ${\gamma }_R$ be positively oriented path on circle of radius $R$ joining endpoints of ${\eta }_{+}$ and ${\eta }_{-}$ on circle of radius $R$

Similarly let ${\gamma }_{ϵ}$ be path on circle of radius $ϵ$ which is negatively oriented and joining endpoints of ${\eta }_{+}$ and ${\eta }_{-}$ on circle of radius $ϵ$

Set

$${\Gamma }_{R,ϵ}={\ell }_{+}\star {\gamma }_R\star {\ell }_{-}\star {\gamma }_{ϵ}$$