Residue theorem

Statement

Let $U \subset \mathbb{C}$ be an open domain, let $S = \{a_1, \dots, a_m\} \subset U$ be a finite set of isolated singularities of a holomorphic function $f : U \setminus S \to \mathbb{C}$, and let $\gamma$ be a positively oriented simple closed contour in $U \setminus S$ whose interior lies in $U$ and contains $S$. Then $\oint_{\gamma} f(z)\,dz = 2\pi i \sum_{k=1}^{m} \operatorname{Res}(f, a_k)$, where $\operatorname{Res}(f, a_k) = c_{-1}$ is the coefficient of $(z - a_k)^{-1}$ in the Laurent series expansion of $f$ around $a_k$.

Why is it true?

By Cauchy's integral theorem, a holomorphic function has zero circulation everywhere, so an integral around a loop only picks up contributions from the 'punctures' $a_k$ inside the loop. Around each puncture, expanding $f(z)$ into a Laurent series $\sum_{n=-\infty}^{\infty} c_n (z - a_k)^n$ shows that every power $(z - a_k)^n$ with $n \ne -1$ has an exact antiderivative $(z - a_k)^{n+1}/(n+1)$ and integrates to zero around a closed loop. Only the $(z - a_k)^{-1}$ term — whose antiderivative is the multi-valued logarithm $\log(z - a_k)$ — fails to cancel, leaving a 'residue' of $2\pi i\,c_{-1}$ per wind around $a_k$. This turns continuous line integrals into pure algebra at a few isolated points.

Topics that use this theorem

Related theorems

Step-by-step proofs

No step-by-step proof yet for this theorem.