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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.

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 $\Omega$.