Power-map reduction for Brownian exit from a wedge ($R=r^{\pi /\alpha }$)

The map $z\mapsto z^{\pi /\alpha }$ takes a wedge of opening $\alpha$ to a half-plane and the circular boundary of radius $r$ to radius $r^{\pi /\alpha }$. Conformal invariance of planar Brownian motion preserves which boundary part is reached first, although it changes the clock. Use the branch defined by the wedge argument; the origin is a polar point for planar Brownian motion, so localization handles the vertex.

For Brownian motion started at $i$ in the upper half-plane, the real-axis hit $T$ is finite and its Cauchy distribution is nonzero almost surely. The continuous path on $[0,T]$ therefore stays in a compact annulus avoiding zero. The derivative of $z^{\alpha }$ is bounded on that annulus, proving finite changed exit time from its wedge image. The exponent zero is excluded because the map is constant and its image is not an open domain.

A planar Brownian motion starting at $1$ exits the disk of radius $r>1$ before leaving a wedge of opening $\alpha \in (0,2\pi ]$, centered on the positive axis, with probability $(4/\pi )\arctan (r^{-\pi /\alpha })$. The power-map reduction for Brownian exit from a wedge, the reflection identity for Brownian exit from a half-disc and the Möbius calculation of circular Brownian exit give the formula. Its large-radius decay exponent is $\pi /\alpha$.