I have the following question about planar brownian motion concerning the following hitting time problem:
Take any $z \in \mathbb H$ with $\Re(z)<0$ and consider the vertical barrier given by $\gamma_t = 2i \sqrt{t}$ growing in time $t$ vertically along the imaginary axis. Consider planar Brownian Motion $(\beta_t)$ starting at $z$ what is the chance that the Brownian motion $\beta$ hits the left side of $\gamma_t$ before the right side of $\gamma_t$. And how does this change asymptotically as $t \to \infty$?
My try: This should be a clean case of the conformal invariance of Brownian motion, the conformal map $g_t : \mathbb H \setminus \gamma_t \to \mathbb H$ given by $g_t(z) = \sqrt{z^2+4t}$ should do the trick, since the tip of the curve $\gamma(t) = 2i\sqrt{t}$ is getting mapped to $0$ while the rest $\gamma([0,t))$ is getting mapped on the real axis. Hence applying the conformal invariance of Brownian motion one should get $$ \mathbb P_{z}\big[\beta \text{ hits left side of } \gamma([0,t]) \text{ or } (-\infty,0) \text{ when exiting }\mathbb H \big] = \mathbb P_{g_t(z)}\big[\beta_{\nu_{\mathbb H}} \in (-\infty, 0) \big], $$ However I am unable to go further on the right side, however my solutions tell me the right side should be given by $\arg(g_t(z))/\pi$, since $\arg(\beta)$ is a martingale (this part is clear to me), however why it should be given by the argument is unclear to me. Does somebody know the answer?
Thanks in advance!