= Power-map reduction for Brownian exit from a wedge
{title2=$R=r^{\pi/\alpha}$}
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.
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