Compute \[ \Res_{z=\infty}e^z, \qquad \Res_{z=\infty}{z-1\over z+1}. \]
Solution
Recall \[ \Res_{z=\infty}f(z) =-\Res_{w=0}\frac{f(1/w)}{w^2}. \]
For \(f(z)=e^z\), \[ \frac{e^{1/w}}{w^2} =\sum_{k=0}^\infty {w^{-k-2}\over k!}. \] There is no \(w^{-1}\) term, hence \[ \Res_{z=\infty}e^z=0. \]
For \[ f(z)={z-1\over z+1}, \] we have \[ \frac{f(1/w)}{w^2} ={1\over w^2}{1-w\over1+w} ={1\over w^2}\qty{1-2w+2w^2-2w^3+\cdots}. \] Thus \[ \Res_{w=0}\frac{f(1/w)}{w^2}=-2, \] so \[ \Res_{z=\infty}{z-1\over z+1}=2. \] Equivalently, the only finite pole is at \(z=-1\) with residue \(-2\), and the sum of all residues on \(\CP^1\) is zero.