The Residue Theorem
Reduce an integral around a closed contour to a sum of residues, one number per singularity enclosed.
What a learner can do afterwards
- 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
1 · Read
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.
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.
Read one coefficient per hole, weight by winding, and scale by 2 pi i.
All nine extra tasks and three readings are original authorship created for this topic.
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Where it sits
17 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.