 Solution

Choose the argument interval $-3\pi /2<\theta <\pi /2$, which makes the logarithm real on the positive real axis and puts its cut on the positive imaginary axis. Parametrize $z=R{e}^{i\theta }$. The constant term $\ln R$ contributes $\ln R\oint dz=0$, leavingEquivalently the cut-crossing point is $iR$, so the circle integral of a logarithm with a radial branch cut is $2\pi i(iR)=-2\pi R$. The different cut changes where the discontinuity lies on the circle; it is not merely an additive constant change of branch along the entire contour.