Residue ($\mathrm{Res}$)

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 asThis 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: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 $g(a)\ne 0$, then $f$ has a simple pole andThe 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. ^[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: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,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. ^[ref]

Why does only this Laurent coefficient survive? Parametrize a counterclockwise circle by $z=a+r{e}^{it}$, $0\le t\le 2\pi$, with $0<r<\rho$. For every integer $n$,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 $\oint f(z)dz=2\pi i\mathrm{Res}(f,a)$. Thus the residue isolates exactly the local information that a closed contour integral detects. ^[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.

At a simple pole, write $f(z)=c/(z-a)+h(z)$ with $h$ holomorphic near $a$ and $c$ its residue. On the arc $z=a+ϵe^{it}$, the singular term integrates exactly to $ic(\beta -\alpha )$. The remainder has absolute integral at most $ϵ(\beta -\alpha )\sup |h|$, which tends to zero. Reversing orientation reverses the sign. In particular a clockwise upper semicircle contributes $-i\pi c$ in the shrinking limit.

A higher-order pole need not have such a limit: $(z-a{)}^{-2}$ has residue zero but integral $2/ϵ$ on a positively oriented upper semicircle. Thus the simple-pole assumption controls all the potentially divergent terms of the principal part of a Laurent series.

If $g,h$ are holomorphic near $a$, $h(a)=0$ and $h^{\prime }(a)\ne 0$, write $h(z)=(z-a)k(z)$ with $k(a)=h^{\prime }(a)$. Then $g/h=(g/k)/(z-a)$, so its residue is $g(a)/h^{\prime }(a)$. If $g(a)\ne 0$ this is a simple pole; if $g(a)=0$ the singularity is removable and the formula gives zero.

Ifwith $g$ holomorphic at $a$, then