 Solution

There is a printed domain error: $\mathrm{Re}z>0$ defines the right half-plane. Taken literally, continuity forces $\mathrm{Re}{Z}_{T_H}=0$ almost surely, so its law is a point mass at zero. The displayed density belongs to the intended upper half-plane $\mathrm{Im}z>0$, with $b>0$. We prove that intended assertion explicitly.

Write $Z_t=a+X_t+i(b+Y_t)$ for independent standard Brownian motions. The exit time $T$ is the first time $Y$ hits $-b$, and depends only on $Y$. Its density, by the supplied Brownian first-passage time law, is $b(2\pi t^3{)}^{-1/2}{e}^{-b^2/(2t)}$. Conditional on $T=t$, the real exit coordinate is normally distributed with mean $a$ and variance $t$. Independence therefore givesWith $u=[b^2+(x-a{)}^2]/(2t)$, the integral is $2/[b^2+(x-a{)}^2]$. Thus the Cauchy exit law from a Brownian half-plane isFor the literal right half-plane, the corresponding nondegenerate coordinate is $\mathrm{Im}{Z}_{T_H}$, whose density is $a/\{\pi [a^2+(y-b{)}^2]\}$ for $a>0$.