Left-semicircle harmonic measure from an off-center point (${\omega }_a(\mathrm{Re}z<0)={\pi }^{-1}\arctan ((1-a^2)/(2a))$)

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.