= Solution
On the unit circle use $z=e^{i\theta}$ with $0<\theta<2\pi$. The specified branch has $\operatorname{Log}z=i\theta$ there and $dz=ie^{i\theta}d\theta$. Therefore
$$
\boxed{I=-\int_0^{2\pi}\theta e^{i\theta}d\theta=-[(1-i\theta)e^{i\theta}]_0^{2\pi}=2\pi i.}
$$
This is the <circle integral of a logarithm with a radial branch cut>, using the one-sided values at the cut crossing. The logarithm is discontinuous at that crossing, so the <Cauchy integral theorem> for a function holomorphic around a closed contour does not apply.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2-logarithm-branch-circles.png]
{title=The cut-crossing point determines the circle integral of a logarithm}
{description=Original unit-circle illustrations of the two radial branch cuts. The formula is $2\pi i$ times the cut-crossing point; the second circle represents the general radius-$R$ case by rescaling.}
Back to article page
