 Solution

Set $\beta =\pi /\alpha$ and use the branch $\varphi (z)=z^{\beta }$ with argument in $(-\alpha /2,\alpha /2)$. This is a conformal map from the wedge to the right half-plane, fixes the starting point $1$, and sends the outer circle of radius $r$ to that of radiusThe two wedge sides map to the imaginary axis.

By conformal invariance of planar Brownian motion, the image of the stopped path is planar Brownian motion after the increasing conformal Brownian clock $A(t)={\int }_0^t|{\varphi }^{\prime }(B_s){|}^{2}ds$. This clock does not change which boundary portion is reached first. Localization away from the vertex justifies the map even when its derivative is unbounded there; the vertex is a polar point for planar Brownian motion and has zero hitting probability from $1$.

Consequentlywhere the probability on the right is for the right half-plane. The power-map reduction for Brownian exit from a wedge also works at $\alpha =2\pi$, when the wedge is the plane slit along the negative real axis.