= Left-semicircle harmonic measure from an off-center point
{title2=$\omega_a(\operatorname{Re}z<0)=\pi^{-1}\arctan((1-a^2)/(2a))$}
For $0<a<1$ in the unit disk, the <Möbius transformation> $(z-a)/(1-az)$ takes $a$ to zero and the left semicircle to the arc centered at $-1$ with half-angle $\arctan((1-a^2)/(2a))$. Brownian exit from the centered disk is uniform in angle, so <conformal invariance of harmonic measure> gives the displayed probability. Scaling and the <reflection identity for Brownian exit from a half-disc> give probability $4\arctan(1/r)/\pi$ of exiting the radius-$r$ circle from start one before reaching the imaginary axis.
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