3.5 Residue at infinity
Suppose \(U\subseteq \mathbb {C}\) is an open subset containing the set \(\{ z: |z| {\gt} r\} \) for some real number \(r\), and \(f\in A(U)\). Let \(r_1 {\gt} r_2 {\gt} r\) and write \(C_i\) for the circle \(|z| = r_i\) oriented counterclockwise. Then
We we write
and call this number the residue of \(f\) at infinity.
The first assertion is easy (and basically already proved in the course of proving the theorem on existence of Laurent series). The point is \(f\) is analytic on the region between the two circles.
Suppose \(U\) and \(f\) are is as in Definition-Proposition 3.29. Then
The open set \(U\) contains the annulus \(A(r) := \{ z\in \mathbb {C}: |z| {\gt} r\} \) for some \(r {\gt} 0\). By the Laurent series theorem, we have a Laurent series \(\sum _{i=-\infty }^{\infty } a_i z^i\) converging to \(f\) on \(A(r)\) with
for any circle \(C\) with center \(0\) contained in \(A(r)\). In particular, \(a_{-1} = -\int _C f(z)\, dz = -\operatorname*{\mathrm{Res}}_{z=\infty } f\). On the other hand, if we set \(g(z) = f(z^{-1})\), we see that \(g(z)\) is analytic on the punctured disk \(D^*(0, 1/r)\) and the power series \(\sum _{i\in \mathbb {Z}} a_i z^{-i}\) converges to \(g(z)\). Then \(g(z)/z^2 = \sum _{k=-\infty }^{\infty } a_k z^{-k-2}\). So \(a_{-1} = \operatorname*{\mathrm{Res}}_{z=0} g(z)/z^2\). Therefore, \(\operatorname*{\mathrm{Res}}_{z=\infty } f(z) = -\operatorname*{\mathrm{Res}}_{z=0} g(z)/z^2\) as desired.