= Solution
{c}
There is a printed domain error: $\operatorname{Re}z>0$ defines the right half-plane. Taken literally, continuity forces $\operatorname{Re}Z_{T_H}=0$ almost surely, so its law is a point mass at zero. The displayed density belongs to the intended upper half-plane $\operatorname{Im}z>0$, 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 gives
$$
h(x)=\frac b{2\pi}\int_0^\infty t^{-2}
\exp\left(-\frac{b^2+(x-a)^2}{2t}\right)dt.
$$
With $u=[b^2+(x-a)^2]/(2t)$, the integral is $2/[b^2+(x-a)^2]$. Thus the <Cauchy exit law from a Brownian half-plane> is
$$
\boxed{h(x)=\frac b{\pi[b^2+(x-a)^2]}.}
$$
For the literal right half-plane, the corresponding nondegenerate coordinate is $\operatorname{Im}Z_{T_H}$, whose density is $a/\{\pi[a^2+(y-b)^2]\}$ for $a>0$.
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