The residue at infinity is minus the $z^{-1}$ coefficient, equivalently the residue at $0$ after inversion.Suggest a slogan
Let \(R\ge0\), let \(f\) be holomorphic on \(\theset{z : \abs{z}>R}\) with Laurent expansion \(f(z)=\sum_{k\in\ZZ}a_kz^k\) there, and let \[ \Res_{z=\infty}f(z) \coloneqq -\frac{1}{2\pi i}\int_{\abs{z}=r}f(z)\dz = -a_{-1} \] for any \(r>R\), with the circle oriented counterclockwise. Then, with \(g(z) \coloneqq -{1 \over z^2}f\qty{1\over z}\) for \(0<\abs{z}<1/R\), \[ \Res_{z=\infty}f(z) = \Res_{z=0} g(z). \]
For \(0<\abs{w}<1/R\), \(g(w)=-w^{-2}\sum_{k\in\ZZ}a_kw^{-k}=-\sum_{k\in\ZZ}a_kw^{-k-2}\). The coefficient of \(w^{-1}\) comes from \(k=-1\) and equals \(-a_{-1}\), so \(\Res_{w=0}g(w)=-a_{-1}=\Res_{z=\infty}f(z)\).