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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?

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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 $\alpha \in \mathbb{C}, \, \alpha \ne 0.$

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  • $\begingroup$ Thanks. Can you explain your symmetry argument for finding $r$? Why should $\varphi(1/2)=r$ and $\varphi(0)=-r$? $\endgroup$ – mathjacks Feb 16 '16 at 19:19
  • $\begingroup$ Also, how did you come up with the map $\varphi(\zeta)$? It seems to come out of nowhere for me. $\endgroup$ – mathjacks Feb 16 '16 at 22:33
  • $\begingroup$ We use several variables $z,\, \zeta,\, \xi,...$, since we consider $\zeta=f(z)=\frac{1}{z},\, \xi=\varphi (\zeta)=\frac{\zeta-c}{1-c\zeta},\, w=\phi(\xi) $ and those composition $\phi\circ \varphi \circ f$. $\endgroup$ – ts375_zk26 Feb 17 '16 at 0:34
  • $\begingroup$ how came up with: --- We want to find $f(z)=\frac{z-a}{1-\bar{a}z} $ (which maps the unit circle onto itself) such that it maps the circle $|z-\frac{1}{4}|=\frac{1}{4}$ onto $|z|=r$. We may specify $f(1)=1$ (if $f(1)=e^{i\theta }$, then consider $e^{-i\theta }f(z)$). Then $\frac{1-a}{1-\bar{a}}=1$ and we see $a$ is real. Thus we consider $\varphi (\zeta)=\frac{\zeta-c}{1-c\zeta}$. $\endgroup$ – ts375_zk26 Feb 17 '16 at 0:34
  • $\begingroup$ symmetry argument: --- $\varphi (\frac{1}{2})$ is a point on $|\xi|=r.$ $\varphi (\frac{1}{2})$ is real, since $\varphi (x)$ is real for real $x$. Thus $\varphi (\frac{1}{2})=r$ or $-r.$ But $\varphi (\frac{1}{2})=-r$ is impossible since $\varphi ([\frac{1}{2},\, 1])$ is a line segment joining $\varphi (\frac{1}{2})$ and $\varphi (1)=1$. $\endgroup$ – ts375_zk26 Feb 17 '16 at 0:36

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