--- title: "The Residue Theorem" description: "Reduce an integral around a closed contour to a sum of residues, one number per singularity enclosed." canonical: https://lightmysky.com/learn/mathematics/the-residue-theorem-mt_BN2biZ_zos source: https://lightmysky.com/learn/mathematics/the-residue-theorem-mt_BN2biZ_zos.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # The Residue Theorem Reduce an integral around a closed contour to a sum of residues, one number per singularity enclosed. Subject: Mathematics · Area: Complex Analysis · Ages 21 to 22 Page: https://lightmysky.com/learn/mathematics/the-residue-theorem-mt_BN2biZ_zos ## Ready when they can - Compute a residue at a simple pole and at a pole of higher order - State the residue theorem with the winding number and apply it to a contour enclosing several poles - Explain why only the coefficient of the minus-one power survives the integration ## Lesson: One coefficient rules each loop The residue of f at an isolated singularity a is the coefficient c_-1 of (z-a)^(-1) in the Laurent expansion on a punctured disk 0 < |z-a| < R. It is a local number depending only on f and a, not on any contour. A zero residue does not remove the pole: 1/(z-a)^2 has residue 0 but is still a pole. At a simple pole the residue is a one step limit: multiply by z minus a and evaluate. For f = g/h with g and h holomorphic near a, h(a) = 0 and h prime(a) nonzero, the residue is g(a) over h prime(a). When g(a) is nonzero this is a genuine simple pole. When g(a) = 0 the factor cancels and the singularity is removable with residue 0. For (z+3)/(z-1) at 1, cover up gives 3+1 = 4. At a pole of order m the function h = (z-a)^m f extends holomorphically to a. The residue is the (m-1)th derivative of h at a, divided by (m-1) factorial. For (z^2+1)/(z-1)^3 at 1, h = z^2+1, its second derivative is 2, so the residue is 2 over 2 factorial = 1. For integer n not equal to minus one, (z-a)^n has primitive (z-a)^(n+1)/(n+1), single-valued on the punctured plane, so its loop integral is zero. For n = minus one no global logarithm exists on the punctured plane. A small circle traversed once counterclockwise integrates (z-a)^(-1) to 2 pi i. For winding 0 the minus-one term also integrates to zero. A Laurent series converges uniformly on any compact circle inside its annulus, which justifies integrating term by term on that circle. Let D be a simply connected open domain and f holomorphic on D except finitely many isolated singularities. Let C be a closed piecewise C1 contour in D avoiding the singularities. Then the integral of f over C equals 2 pi i times the sum of n(C, a_k) times Res(f, a_k) over the singularities a_k. The winding number n(C, a) counts signed turns: counterclockwise positive, clockwise negative. For a simple closed positively oriented contour with f holomorphic inside except finitely many isolated singularities, each enclosed winding is 1. **Example.** Worked contour application. Integrate (2z+1)/(z(z-3)) around the circle |z| = 2 traversed once counterclockwise. Poles sit at 0 and 3; only 0 lies inside. The residue at 0 is (2*0+1)/(0-3) = -1/3. The integral is 2 pi i times (-1/3) = -2 pi i over 3. Orientation reversal negates every winding number and the integral, but never changes the residues. The residue theorem applies to essential singularities too. For e^(1/z) at 0 the Laurent series is 1 + 1/z + 1/(2z^2) + ..., so the residue is 1. The higher-pole derivative formula does not apply to an essential singularity. **Recap.** Read one coefficient per hole, weight by winding, and scale by 2 pi i. ## Practice 17 questions on this page, each with its working shown. ## Needs first - [Laurent Series and Classifying Isolated Singularities](https://lightmysky.com/learn/mathematics/laurent-series-and-classifying-isolated-singularities-mt_WoQV0gD1O-) ## Opens up - [The Argument Principle and Rouche's Theorem](https://lightmysky.com/learn/mathematics/the-argument-principle-and-rouches-theorem-mt_J1dC428M84) - [Evaluating Real Integrals by Residues](https://lightmysky.com/learn/mathematics/evaluating-real-integrals-by-residues-mt_Lj_HXtjzAh)