Contour Integrals Along Parametrised Paths
Integrate a complex function along a path by parametrising it, and bound the result by length times maximum modulus.
What a learner can do afterwards
- Evaluate a contour integral directly from a parametrisation using complex multiplication
- Show the value is unchanged by orientation-preserving reparametrisation with the same traversal, negated by reversal, and additive over concatenated paths
- Apply the estimation lemma to bound an integral without evaluating it, using any valid modulus bound and total path length
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A contour integral accumulates f along a parametrised path. Write γ(t) = x(t) + i y(t) for t in [a, b]. Then ∫_γ f(z) dz = ∫_a^b f(γ(t)) γ'(t) dt. The integrand and velocity are complex numbers, so their product uses complex multiplication, not a dot product. The derivative is in the real parameter t. The integrand need not be complex differentiable. For f(z) = 1, the integral is the endpoint minus the start point.
Assume f is continuous on the path and γ is piecewise continuously differentiable on [a, b] with a < b. This means the parameter interval splits into finitely many subintervals where γ has a continuous derivative in the real parameter t. Complex differentiability of f is not required.
Worked slanted path. Let γ(t) = 1 + t + it for t in [0, 1]. Then γ'(t) = 1 + i. For f(z) = z, the integral is ∫_0^1 (1 + t + it)(1 + i) dt. Expand: (1 + t + it)(1 + i) = 1 + i + 2it. Integrate componentwise: real part ∫_0^1 1 dt = 1, imaginary part ∫_0^1 (1 + 2t) dt = 2. Result: 1 + 2i.
Circles use radians. Let γ(t) = e^(it) = cos(t) + i sin(t) for t in [0, 2π], traversing once anticlockwise. Then γ'(t) = i e^(it). For f(z) = 1/z, the integrand is 1/γ(t) = e^(-it), and the product is e^(-it) · i e^(it) = i. So ∫_γ 1/z dz = ∫_0^(2π) i dt = 2πi.
A direction-preserving reparametrisation with the same oriented traversal and multiplicity leaves the integral unchanged. For example, γ(t) on [0, 2] becomes γ(2s) on [0, 1]. Reversing a path negates the integral. Zero stays zero under reversal. Adding pieces: ∫_(γ1+γ2) f dz = ∫_γ1 f dz + ∫_γ2 f dz. Path length counts repeated traversal.
The estimation lemma bounds without evaluating. If |f(z)| ≤ M on γ and γ has total length L, then |∫_γ f(z) dz| ≤ M L. M may be any valid bound, not necessarily an attained maximum. For two pieces, |I1 + I2| ≤ |I1| + |I2| ≤ M1 L1 + M2 L2.
For a circle of radius 2, the length is 4π. Using π < 3.15 gives length < 12.6. If |f| ≤ 3, the bound is 3 × 12.6 = 37.8. A rounded-down circumference like 6.28 for 2π is below the true length and cannot guarantee an upper bound.
Feed the path into the function, weight by velocity, and preserve the same oriented traversal.
All nine extra tasks and three readings are original authorship created for this topic.
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