Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2009 / B / 4 / c / Solution

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 $r>1$.