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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.

On the unit circle use $z=e^{i\theta }$ with $0<\theta <2\pi$. The specified branch has $\mathrm{Log}z=i\theta$ there and $dz=i{e}^{i\theta }d\theta$. ThereforeThis 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.