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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 $\pi /\alpha$.

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.

Let $E_{+}$ and $E_{-}$ denote positive and negative real part at the circular exit. On $\{S<T(r)\}$, reflect the portion of the planar Brownian motion after $S$ by $z\mapsto -\overline{z}$. This reflection fixes the imaginary axis and preserves distances from the origin. The Strong Markov property and reflection symmetry show that the resulting path has the same law, its circular exit time is unchanged, and $E_{+}$ is exchanged with $E_{-}$. ThereforeAn exit with negative real part must first cross the imaginary axis. An exit before $S$ has positive real part. The two points $ir,-ir$ have zero circular exit probability, since circular harmonic measure has no atoms. HenceSubtracting provesThis is the reflection identity for Brownian exit from a half-disc.

Continuity and the regularity of zero for a one-dimensional Brownian coordinate make $S$ the first contact with the imaginary axis. The Strong Markov property and symmetry of the subsequent real increments imply that reversing those increments after $S$ preserves the Brownian law. The imaginary coordinate is unchanged, so $\tilde{Z}$ is again planar Brownian motion started at one.

Reflection preserves distance to the origin, hence ${\tilde{T}}_{D_r}=T_{D_r}$ pathwise. On $T_{D_r}<S$, the path exits through the right semicircle and reflection has not yet acted. On $S<T_{D_r}$, reflection reverses the sign of the exit real part. ThusThe two circular points on the imaginary axis have zero exit probability, so they cause no extra term. Subtracting proves the reflection identity for Brownian exit from a half-disc requested in the question.

Both processes have the same exit law. With $p={\pi }^{-1}{\tan }^{-1}((r^2-1)/(2r))$ from part (b),The final trigonometric identity follows from ${\tan }^{-1}((r^2-1)/(2r))=\pi /2-2{\tan }^{-1}(1/r)$ for $r>1$.