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For an inner circle of real center $c$ and radius $d$ lying strictly inside the unit circle, choose the real root $|a|<1$ of $c{a}^2-(1+c^2-d^2)a+c=0$. When $c=0$, take $a=0$. The automorphism of the unit disk $T_a$ sends the inner circle to a concentric circle. Indeed it sends the two real endpoints $c-d,c+d$ to opposite real numbers; preservation of circles and reflection symmetry then give an image centered at zero. Its radius is $r=T_a(c+d)\in (0,1)$.

The conformal invariance of harmonic measure and harmonic measure of an annulus give outer-boundary measure $\log (|T_a(z)|/r)/\log (1/r)$. For $c=d=2/5$, the quadratic gives $a=1/2$ and $r=1/2$, an explicit eccentric-annulus example.

For $0<a<1$ in the unit disk, the Möbius transformation $(z-a)/(1-az)$ takes $a$ to zero and the left semicircle to the arc centered at $-1$ with half-angle $\arctan ((1-a^2)/(2a))$. Brownian exit from the centered disk is uniform in angle, so conformal invariance of harmonic measure gives the displayed probability. Scaling and the reflection identity for Brownian exit from a half-disc give probability $4\arctan (1/r)/\pi$ of exiting the radius-$r$ circle from start one before reaching the imaginary axis.

The concentric normalization of an eccentric annulus here usesThis Möbius transformation preserves the unit circle and sends the inner circle's real endpoints $0,4/5$ to $-1/2,1/2$. Its image circle is symmetric about the real axis and hence is $|w|=1/2$. Equivalently, expanding $|T(z){|}^2=1/4$ gives $|z{|}^2=(4/5)\mathrm{Re}z$, precisely the inner-circle equation. Thus $T$ sends the domain to $1/2<|w|<1$. Conformal invariance of harmonic measure and i giveThe zero of $T$ lies in the removed inner disk, so this logarithm is harmonic throughout $\Omega$.

Let $a=1/r\in (0,1)$ and $\theta ={\tan }^{-1}((1-a^2)/(2a))\in (0,\pi /2)$. The Möbius transformation sends the semicircle endpoints toAlso $f(-1)=-1$. Therefore the left semicircle maps to the unit-circle arc centered at $-1$ with angular width $2\theta$.

By conformal invariance of harmonic measure, its exit probability from the starting point $a$ equals the probability of this image arc for Brownian motion started at zero. Rotational symmetry makes the latter exit angle uniform. ThusFor scaling, if $Z$ starts at $1/r$, then ${\hat{Z}}_t=r{Z}_{t/r^2}$ is planar Brownian motion started at one. Its unit-disk exit becomes radius-$r$ exit, and multiplication by the positive number $r$ preserves the sign of the real part. This proves exactly the same formula for ${\mathbb{P}}_1(\mathrm{Re}{\hat{Z}}_{T_{D_r}}<0)$.

For $z=x+iy$ in the strip,so $\mathrm{Im}f(z)>0$. This is a bijective conformal map onto the upper half-plane. One way to verify bijectivity is to factor it as $w=e^{\pi z/a}$, followed by $(w+w^{-1})/2$. The first map gives $\mathrm{Im}w>0$, $|w|>1$. The two inverse roots of the second map have product one; for any point in the upper half-plane exactly one root has modulus greater than one, and its imaginary part is positive because $\mathrm{Im}[(w+w^{-1})/2]=(1-|w{|}^{-2})\mathrm{Im}w/2$.

The three boundary pieces map as follows: the lower edge goes to $(1,\infty )$, the upper edge to $(-\infty ,-1)$, and the vertical edge to $(-1,1)$. Thus the desired exit event becomes exit from the upper half-plane to the left of $-1$. WriteBy conformal invariance of harmonic measure and the Cauchy exit law from a Brownian half-plane,The strip exit is finite almost surely, since its imaginary coordinate exits $(0,a)$ in finite time; corners have probability zero because individual points are polar points for planar Brownian motion. This is the Top-edge exit probability of a Brownian half-strip.