What is the form of the general mobius transformations that map the line $\Re(z)=2$ and the circle $|z|=1$ into concentric circles centered at $a$?
I'm practicing with mobius transformations and have trouble on this question. I've tried mapping the line and the circle using $f(z) = 1/z$ and then combining the images using some sort of translation but that didn't get anywhere.
In order to find the "most general" family of such mobius transformations, I think I need to use $f(z) = \frac{z-a}{\overline a z - 1}$ which maps the unit circle onto itself, but how should this be incorporated?
First $\zeta=f(z)=\frac{1}{z}$ maps the line $\Re z=2$ into $|\zeta-\frac{1}{4}|=\frac{1}{4}$ and the circle $|z|=1$ into $|\zeta|=1$.
Now we consider $\xi=\varphi (\zeta )=\frac{\zeta -c}{1-c\zeta }, \, 0<c<1.$
Of course it's aime is to map two circles $|\zeta-\frac{1}{4}|=\frac{1}{4}$ and $|\zeta|=1$ into concentric circles $|\xi |=r$ and $|\xi |=1$ centered at $0$. ($r$ is unknown at present.)
We determine $r$. By the symmetry of two circles $|\zeta-\frac{1}{4}|=\frac{1}{4}$ and $|\zeta|=1$ with respect to the real axis we assume $\varphi (\frac{1}{2})=r$ and $\varphi (0)=-r$. Thus $$ \frac{1-2c}{2-c}=r,\quad -c=-r.$$ Hence we get $c=r=2-\sqrt{3}$.
Finally take $w=\phi(\xi)=\alpha \xi +a$ or $w=\phi(\xi)=\frac{\alpha }{\xi }+a$, where $\alpha $ is a non-zero arbitrary complex number. The choice of $\alpha $ is arbitrary and it guarantees the generality of mobius transformations we consider.
Thus the form of the general mobius transformations that map the line $\Re(z)=2$ and the circle $|z|=1$ into concentric circles centered at $a$ is $$ w=(\phi\circ\varphi\circ f)(z)=\alpha\cdot \frac{1-(2-\sqrt{3})z}{z-(2-\sqrt{3})}+a $$ and $$ w=\alpha\cdot \frac{z-(2-\sqrt{3})}{1-(2-\sqrt{3})z}+a, $$ where $\alpha \in \mathbb{C}, \, \alpha \ne 0.$