= 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 $z=Re^{i\theta}$. The constant term $\ln R$ contributes $\ln R\oint dz=0$, leaving
$$
I=-R\int_{-3\pi/2}^{\pi/2}\theta e^{i\theta}d\theta=\boxed{-2\pi R.}
$$
Equivalently 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.
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