Above is my question. The issue that I'm having is that I find that the given mapping, call it $f$, maps $C_1$ to itself, and $C_2$ maps to a circle of radius $5$ and centre $-2$. As much as this in itself isn't a problem, it doesn't really help me with my question!
We have $f(1) = -1,$ $f(-1) = 1$ and $C_1$ perpendicular to $\Bbb R$ at $1$, so this gives $f(C_1) = C_1$; similarly for $C_2$. Also, $f$ is self-inverse.
How can I use the mapping to solve the equation?
Thanks in advance!
Wrong. $f(C_2)$ is the circle of radius $2$ centered at $0$. Justification below.
Observe that $C_2$ is symmetric about the real axis. It crosses the axes at $0$ and $4/5$. The map $f$ preserves the real axis. Also, $f(0)=2$ and $f(4/5)=-2$. Hence, $[-2,2]$ is a diameter of $f(C_2)$.
Now that you have concentric circles, it remains to fit a function of the form $a\log|z|+b$ to the boundary values, and then return to original domain.