Residue (source code)
= Residue
{title2=$\operatorname{Res}$}
{wiki=Residue_(complex_analysis)}
The <residue> of a <function> $f$ at an <isolated singularity> $a$ is its <Laurent coefficient> of degree $-1$. More explicitly, if $f$ is <holomorphic> for $0<|z-a|<\rho$, write its <Laurent series> as
$$
f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n,
\qquad \operatorname{Res}(f,a)=c_{-1}.
$$
This one <Laurent coefficient> determines the <contour integral> around a sufficiently small counterclockwise <circle> about $a$. A complicated local <Laurent series> can therefore leave a remarkably simple answer after one complete trip around the point.
For a <simple pole>, multiplying by $z-a$ removes the <pole>, leaving a <holomorphic> extension whose value at $a$ is the <residue>:
$$
\operatorname{Res}(f,a)=\lim_{z\to a}(z-a)f(z).
$$
There is a useful shortcut for a quotient $f=g/h$. If $g$ and $h$ are <holomorphic> near $a$, $h$ has a <simple zero> there, and $g(a)\ne0$, then $f$ has a <simple pole> and
$$
\operatorname{Res}(g/h,a)=\frac{g(a)}{h'(a)}.
$$
The first-order <Taylor series> of $h$ explains the denominator: dividing $h(z)$ by $z-a$ leaves its <derivative> at $a$ in the limit. This computes a <residue> without finding the full <Laurent series>, which is especially useful when the numerator or denominator is complicated. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/3c58ce71ec6f7573804221d76158a0bd_MIT18_04S18_topic8.pdf[Orloff, MIT Topic 8, section 8.4.1]{ref}
For example, the <rational function> $f(z)=1/(z^2+1)$ has <simple poles> at $i$ and $-i$. Factoring the denominator makes both <residues> immediately visible:
$$
f(z)=\frac{1}{(z-i)(z+i)},\qquad
\operatorname{Res}(f,i)=\frac{1}{2i}=-\frac{i}{2},\qquad
\operatorname{Res}(f,-i)=\frac{1}{-2i}=\frac{i}{2}.
$$
The signs matter: these <residues> cancel when both <poles> are enclosed, but either one contributes when enclosed alone. The <residue theorem> below turns this distinction into different <contour integrals> of the same <function>.
A <residue> is not the same thing as a <pole>, nor does it measure the order of a <pole>. The <function> $1/z^2$ has a <double pole> at zero but <residue> zero. Conversely,
$$
e^{1/z}=1+\frac1z+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots
$$
has an <essential singularity> at zero and <residue> one. Infinitely many negative powers cause no difficulty in reading off that particular <Laurent coefficient>. At a <removable singularity> the <residue> is zero. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/dff6a0c70eefb1e23bb87f8524361801_MIT18_04S18_topic7.pdf[Orloff, MIT Topic 7, sections 7.9–7.9.2]{ref}
Why does only this <Laurent coefficient> survive? Parametrize a counterclockwise <circle> by $z=a+re^{it}$, $0\leq t\leq2\pi$, with $0<r<\rho$. For every <integer> $n$,
$$
\oint_{|z-a|=r}(z-a)^n\,dz
=ir^{n+1}\int_0^{2\pi}e^{i(n+1)t}\,dt
=\begin{cases}2\pi i,&n=-1,\\0,&n\ne-1.\end{cases}
$$
When $n\ne-1$, the <complex exponential> completes a whole number of turns and its contributions cancel; when $n=-1$, it is constant. <Uniform convergence> of the <Laurent series> on this <circle> justifies <termwise integration>, giving $\oint f(z)\,dz=2\pi i\operatorname{Res}(f,a)$. Thus the <residue> isolates exactly the local information that a closed <contour integral> detects. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/dff6a0c70eefb1e23bb87f8524361801_MIT18_04S18_topic7.pdf[Orloff, MIT Topic 7, sections 7.7, 7.9.2 and 7.10]{ref}
This selection rule explains why a very large value near an <isolated singularity> need not produce a large <contour integral>. The <double pole> $1/z^2$ grows faster near zero than $1/z$, yet its contributions around a <circle> cancel completely. What matters is the arrangement of complex values along the path, including the differential $dz$, rather than their <moduli> alone. A <residue> records that surviving contribution, not the size of the <function> near the point.
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