 Solution

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 isFor 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 $(2\pi n{)}^{-1/2}$, henceThus 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 $\mathbb{E}{Y}_{T_S}=Y_0=1$. Since the exit coordinate is zero or $r$,If $T$ is the first hit of the real axis, exit through the lower boundary implies $T<\infty$, so $\mathbb{P}(T<\infty )\ge 1-1/r$ for every $r>1$. Letting $r\to \infty$ provesThe 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.