I've been trying to solve the following problem:
Find a function $\varphi$ which is
- harmonic in the upper half-plane exterior to the circle $|z-5i| = 4$,
- is $1$ on $|z-5i| = 4$,
- and $0$ on the real axis.
I anticipate that the function will be the imaginary part of a conformal mapping.
By a composition of two Mobius transformations it is possible to map $Im(z) = 0$ to $|z| = 1$ and the circle $|z-5i| = 4$ to $|z| = a$ for some $a < 1$, so that the region is mapped to the annulus $a < |z| < 1$.
From here, I'm not sure where to proceed (if this is the right direction), other than the fact that it is straightforward to find a radial harmonic function $v(r)$ for which $v(a) = 1$ and $v(1) = 0$.
Any other suggestions are appreciated!
The function $v(w)=\log |w|/\log a$ is harmonic in the annulus $a<|w|<1$ and $v=0$ on $|w|=1$, $v=1$ on $ |w|=a$ as easily verified.
Thus, if $w=f(z)$ is a Mobius transformation which maps the region to the annulus $a<|w|<1$ (Im $z=0$ be mapped to $|w|=1$), $\varphi (z)=v(f(z))$ is the one required.
By the symmetry, $f(z)$ will be the form $f(z)=(z-ip)/(z+ip)$, where $p$ is real and positive. Again by the symmetry,$f(i)=-a$ and $f(9i)=a$. Thus, we have $p=3$ and $a=1/2$.
The answer of the problem is $\varphi (z)=\log \vert\frac{z+3i}{z-3i}\vert/\log 2$.