Reflection identity for Brownian exit from a half-disc ($\mathbb{P}(T<S)=\mathbb{P}(\mathrm{Re}{B}_T>0)-\mathbb{P}(\mathrm{Re}{B}_T<0)$)

For planar Brownian motion starting on the positive real axis inside a disk, let $T$ be circular exit and $S$ first contact with the imaginary axis. Reflecting the path after $S$ fixes the disk and swaps positive and negative circular exits. The Strong Markov property makes their probabilities equal on $S<T$. Therefore $\mathbb{P}(T<S)=\mathbb{P}(\mathrm{Re}{B}_T>0)-\mathbb{P}(\mathrm{Re}{B}_T<0)$.