Residue theorem
Theorem. Let \(f\) be holomorphic on \(U\) except for finitely many isolated singularities \(a_1,\dots,a_N\), and let \(\gamma\) be a closed contour avoiding them. Then \[ \int_\gamma f(z)\,\mathrm{d}z=2\pi i\sum_{j=1}^N n(\gamma,a_j)\operatorname{Res}(f,a_j), \] where \(n(\gamma,a_j)\) is the winding number.
Simple pole residue
Proposition. If \(f=g/h\), where \(g,h\) are holomorphic, \(h(a)=0\), and \(h'(a)\ne0\), then \(a\) is a simple pole and \(\operatorname{Res}(f,a)=g(a)/h'(a)\).
Higher-order pole formula
Proposition. If \(f\) has a pole of order \(m\) at \(a\), then \[ \operatorname{Res}(f,a)=\frac{1}{(m-1)!}\left[\frac{\mathrm{d}^{m-1}}{\mathrm{d}z^{m-1}}(z-a)^m f(z)\right]_{z=a}. \]
Proof by puncturing
Proof. Remove small circles around each singularity. Cauchy's theorem applies on the punctured region. Each small boundary circle contributes \(2\pi i\) times the local \((z-a_j)^{-1}\) coefficient, with sign determined by orientation and winding number.
Residue at infinity
Definition. For a meromorphic function near infinity, define \(\operatorname{Res}(f,\infty)=-\operatorname{Res}(z^{-2}f(1/z),0)\). On the Riemann sphere, the sum of all residues, including infinity, is zero.
Computational principle
Remark. Residues convert global contour integrals into local coefficient computations. The art is choosing a contour whose extra pieces vanish or combine into the desired integral.
Residues as local contour invariants
Definition. For an isolated singularity \(a\), the residue is the Laurent coefficient \[ \operatorname{Res}(f,a)=c_{-1} =\frac1{2\pi i}\int_{|z-a|=\rho}f(z)\,\mathrm{d}z. \]
Theorem (residue theorem). Let \(\Gamma\) be a cycle null-homologous in \(\Omega\), and let \(f\) be meromorphic inside it with finitely many poles \(a_j\) off \(\Gamma\). Then \[ \int_\Gamma f(z)\,\mathrm{d}z =2\pi i\sum_j \operatorname{Ind}(\Gamma,a_j)\operatorname{Res}(f,a_j). \]
Proof. Subtract the principal parts needed only at order \((z-a_j)^{-1}\) for integration: \[ g(z)=f(z)-\sum_j\frac{\operatorname{Res}(f,a_j)}{z-a_j}. \] Terms of order other than \(-1\) have zero small-circle integral, and after subtracting complete principal parts one obtains a holomorphic function on the punctures. Apply Cauchy's theorem and use \(\int_\Gamma(z-a_j)^{-1}\mathrm{d}z =2\pi i\,\operatorname{Ind}(\Gamma,a_j)\).
Computing residues
- At a simple pole of \(g/h\), with \(h(a)=0\), \(h'(a)\ne0\), \[ \operatorname{Res}\left(\frac gh,a\right)=\frac{g(a)}{h'(a)}. \]
- At a pole of order \(m\), \[ \operatorname{Res}(f,a) =\frac1{(m-1)!} \left. \frac{\mathrm{d}^{m-1}}{\mathrm{d}z^{m-1}} \{(z-a)^mf(z)\}\right|_{z=a}. \]
- If \(f=g'/g\), the residue at a zero or pole of \(g\) is its signed order.
Residue at infinity
Proposition. For a meromorphic function on the Riemann sphere, \[ \operatorname{Res}(f,\infty) =-\operatorname{Res}_{w=0}\frac{f(1/w)}{w^2}, \] and \[ \sum_{a\in\mathbb C}\operatorname{Res}(f,a) +\operatorname{Res}(f,\infty)=0. \]
Proof. Substitute \(z=1/w\) in a large positively oriented circle; orientation reverses and \(\mathrm{d}z=-w^{-2}\mathrm{d}w\). The residue theorem on the sphere, whose compact surface has no boundary, gives the sum-zero identity.
For a rational function, the residue at infinity is often the fastest way to obtain the sum of all finite residues.