---
title: "Contour Integrals Along Parametrised Paths"
description: "Integrate a complex function along a path by parametrising it, and bound the result by length times maximum modulus."
canonical: https://lightmysky.com/learn/mathematics/contour-integrals-along-parametrised-paths-mt_CaJ7ruAv8P
source: https://lightmysky.com/learn/mathematics/contour-integrals-along-parametrised-paths-mt_CaJ7ruAv8P.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`.

# Contour Integrals Along Parametrised Paths

Integrate a complex function along a path by parametrising it, and bound the result by length times maximum modulus.

Subject: Mathematics · Area: Complex Analysis · Ages 20 to 21
Page: https://lightmysky.com/learn/mathematics/contour-integrals-along-parametrised-paths-mt_CaJ7ruAv8P

## Ready when they can

- 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

## Lesson: Walking a function along a path

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.

**Example.** 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.

**Example.** 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.

**Recap.** Feed the path into the function, weight by velocity, and preserve the same oriented traversal.

## Practice

17 questions on this page, each with its working shown.

## Needs first

- [Complex Differentiability and the Cauchy-Riemann Equations](https://lightmysky.com/learn/mathematics/complex-differentiability-and-the-cauchy-riemann-equations-mt_nZNmZcl9Ss)
- [Vector Fields and Line Integrals](https://lightmysky.com/learn/mathematics/vector-fields-and-line-integrals-mt_VZxKEmsVgO)

## Optional connections

- [Harmonic Functions and What Analyticity Forces](https://lightmysky.com/learn/mathematics/harmonic-functions-and-what-analyticity-forces-mt_jqajP39XI4)

## Opens up

- [Cauchy's Theorem and Deforming a Contour](https://lightmysky.com/learn/mathematics/cauchys-theorem-and-deforming-a-contour-mt_aacYFVN-HH)