---
title: "Define complex integral; evaluate ∮log z dz"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:complex-integral-log-z"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/complex-integral-log-z
site: BCT Engineering
---

# Define complex integral; evaluate ∮log z dz

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2070 Magh, 5 marks
- Video: [Complex Analysis integration of log z along |z| =1 || Complex Integral bsc sem 5 maths || be meths](https://www.youtube.com/watch?v=_DfyfHcD10k) — Swati Theng Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/complex-integral-log-z
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Define complex integral. Evaluate ∮_C log z dz; C: |z| = 1.

## Short answer

The complex integral ∫_ C f(z)dz = ∫_ C (u dx − v dy) + i∫_ C (v dx + u dy). For C:|z|=1, set z=e^{iθ}, principal branch log z = iθ, dz = i e^{iθ} dθ, θ∈(−π,π). Then ∮ log z dz = −∫_{−π}^{π} θ e^{iθ} dθ; odd part θ cosθ gives zero, even part θ sinθ gives 2π, yielding −2πi.

## Answer

#### Complex integral
For a continuous f(z) on a curve C, the **complex integral** is

∫_C f(z) dz = ∫_C (u dx − v dy) + i ∫_C (v dx + u dy) = ∫ f(z(t)) z′(t) dt.

#### Solution
C: |z| = 1, so z = e^{iθ} and dz = i e^{iθ} dθ. Take the principal branch, log z = ln|z| + i Arg z = iθ with −π < θ < π. (log z is not analytic inside C because of the branch point at 0 and the cut, so Cauchy's theorem cannot be used.)

```text
I = ∫_{−π}^{π} (iθ)(i e^{iθ}) dθ
  = − ∫_{−π}^{π} θ e^{iθ} dθ
  = − ∫ θ cos θ dθ − i ∫ θ sin θ dθ
```

The first integral is odd (θ cos θ is odd) so it is 0 over (−π, π). The second is even:

∫_{−π}^{π} θ sin θ dθ = 2 [ −θ cos θ + sin θ ]₀^π = 2 (π) = 2π

So I = −i · 2π = −2πi.

**Answer: ∮_C log z dz = −2πi** (principal branch, counter-clockwise, starting at θ = −π).

If the branch cut is taken along the positive real axis instead (0 ≤ θ < 2π), the same working gives +2πi, so the value depends on the branch chosen.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why can Cauchy's theorem not be used to show the integral of log z around |z|=1 is zero?
   A. log z is not analytic inside C because of the branch point at 0 and the cut
   B. The contour |z|=1 is not a closed curve
   C. log z has a simple pole at z = 0 with a non-zero residue
   D. log z is not continuous anywhere on a circle
   Answer: A. The principal log z has a branch point at 0 and a cut, so it is not analytic inside C and the theorem does not apply. (Key point: Why Cauchy fails)

2. On |z|=1 with the principal branch, what are log z and dz in terms of theta?
   A. log z = θ, dz = e^{iθ} dθ
   B. log z = iθ, dz = i e^{iθ} dθ
   C. log z = ln θ, dz = i dθ
   D. log z = iθ, dz = e^{iθ} dθ
   Answer: B. With z = e^{iθ}, ln|z| = 0 so log z = iθ, and dz = i e^{iθ} dθ. (Key point: Parametrisation)

3. After substitution, the integral becomes -∫θ e^{iθ} dθ over (-π, π). Which part survives?
   A. Only the real part, from θ cos θ
   B. Both parts, each equal to π
   C. Only the imaginary part, from the even function θ sin θ
   D. Neither; both vanish by oddness
   Answer: C. θ cos θ is odd and integrates to 0, while θ sin θ is even and gives 2π, so I = -i(2π). (Key point: Reduction to real integrals)

4. What is the value with the principal branch, and what happens if the cut is on the positive real axis (0 ≤ θ < 2π)?
   A. 0; the value is independent of branch
   B. 2πi; the other branch gives -2πi
   C. -2πi; the other branch also gives -2πi
   D. -2πi; the same working gives +2πi, so it depends on the branch
   Answer: D. The principal branch gives -2πi and the 0 to 2π branch gives +2πi, so the value depends on the branch. (Key point: Final value and branch)

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