Möbius calculation of circular Brownian exit (${\psi }_a(w)=(w-a)/(1-aw)$)

A planar Brownian motion beginning at $a\in (0,1)$ in the unit disc has circular exit law obtained by pulling back uniform angular measure through ${\psi }_a(w)=(w-a)/(1-aw)$. This Möbius transformation sends the starting point to zero. The right semicircle maps to an arc with endpoints at angles $\pm (\pi /2+2\arctan a)$, giving positive-half-plane exit probability $1/2+(2/\pi )\arctan a$.