Contour integration

Objective. Parametrise and evaluate contour integrals.

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Learn: the key idea

Contour integration extends integration to paths in the complex plane: for a curve C parametrised by z(t), a ≤ t ≤ b, ∫_C f(z) dz = ∫ₐᵇ f(z(t)) z'(t) dt. The value can depend on the path unless f is holomorphic on a simply connected region containing it, in which case Cauchy's theorem makes the integral path-independent and often zero around closed loops. For f(z) = z along the unit circle z(t) = e^{it}, 0 ≤ t ≤ 2π, ∫_C z dz = ∫₀^{2π} e^{it}·ie^{it} dt = i∫₀^{2π} e^{2it} dt = 0, since e^{2it} integrates to zero over a full period.

Worked example

Evaluate the contour integral of f(z) = 1/z around the unit circle |z| = 1, traversed counterclockwise.

1 $Parametrise the contour as z(t) = e^{it}, 0 ≤ t ≤ 2π, so z'(t) = ie^{it}.$
2 $Compute the integrand: f(z(t))z'(t) = (1/e^{it})·ie^{it} = i.$
3 $Integrate the constant i over the parameter range: ∫₀^{2π} i dt = 2πi.$
4 $So the contour integral of 1/z around the unit circle equals 2πi.$

Answer: 2πi

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Common mistake. Students forget to multiply the integrand by z'(t) when converting a contour integral to a real parameter integral, omitting the velocity factor and producing an incorrect result.

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