= Solution
{c}
Write $Z=X+iY$, with $Y$ a real <Brownian motion> started at one. For the nontrivial strip calculation assume $r>1$, so the starting point is in its interior. The first exit time is
$$
T_S=\inf\{t\geq0:Z_t\notin S\}.
$$
For a continuous Brownian path started inside, this equals the first hitting time of $Y=0$ or $Y=r$ almost surely: <Brownian motion> started at a boundary immediately visits the exterior, so using the closed strip in the definition does not change the infimum.
If $T_S>n$, then $Y_n\in[0,r]$. Its normal density is bounded by $(2\pi n)^{-1/2}$, hence
$$
\mathbb P(T_S>n)\leq\mathbb P(0\leq Y_n\leq r)
\leq\frac r{\sqrt{2\pi n}}\longrightarrow0.
$$
Thus the exit time is finite almost surely.
The stopped harmonic coordinate $Y_{t\wedge T_S}$ is a <martingale> bounded between zero and $r$. <bounded convergence theorem> and <optional stopping theorem> give $\mathbb EY_{T_S}=Y_0=1$. Since the exit coordinate is zero or $r$,
$$
\boxed{\mathbb P(Y_{T_S}=r)=\frac1r,\qquad
\mathbb P(Y_{T_S}=0)=1-\frac1r.}
$$
If $T$ is the first hit of the real axis, exit through the lower boundary implies $T<\infty$, so $\mathbb P(T<\infty)\geq1-1/r$ for every $r>1$. Letting $r\to\infty$ proves
$$
\boxed{\mathbb P(T<\infty)=1.}
$$
The strip assumption is worth making explicit: at $r=1$ the initial point is the upper boundary and exit is immediate; for $0<r<1$ it starts outside the strip and the displayed interior probabilities do not apply.
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