Residue at infinity

$\operatorname{Res}_{z=\infty} f := -\operatorname{Res}_{w=0}\left[\frac{1}{w^2}f\!\left(\frac1w\right)\right]$ , chosen so all residues, including infinity's, sum to zero.

The residue at infinity of $f$ is defined by substituting $w=1/z$ and taking the residue of the resulting function at $w=0$ : $\operatorname{Res}_{z=\infty} f := -\operatorname{Res}_{w=0}\left[\dfrac{1}{w^2}f\!\left(\dfrac1w\right)\right]$ , the minus sign coming from the orientation reversal that substitution causes. It measures $f$ 's behavior as $|z|\to\infty$ the same way an ordinary residue measures behavior near a finite point, and if $f$ has only finitely many singularities in the plane, the sum of all its finite residues together with its residue at infinity is always exactly zero.

Examples

For $f(z)=1/z$ , substituting gives $\frac1{w^2}\cdot w = \frac1w$ , so $\operatorname{Res}_{z=\infty}\frac1z = -1$ .
Because residues (finite and infinite) sum to zero, computing every finite residue of a rational function and subtracting their sum gives the residue at infinity without any substitution at all.