Define complex integral; evaluate ∮log z dz

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.)

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.