Mapping $|z|=1$ and $|z-\frac{1}{4}|=\frac{1}{4}$ into concentric circles using the cross ratio

Question

The question: Find a linear transformation which carries $|z|=1$ and $|z-\frac{1}{4}|=\frac{1}{4}$ into concentric circles and find the ratio of the radii.
This question is already answered here: Find the linear fractional transformation that maps the circles |z-1/4| = 1/4 and |z|=1 onto two concentric circles centered at w=0?

However, I was hoping to be able to set this problem up straight from the beginning using the cross ratio. So, from the above solution, I can use, where $w=f(z)$, $f(1)=1$ and $f(-1)=-1$. But I am a bit confused on how to handle the circles. For instance, if I was given a circle and then given the circle that it is mapped into, I could find a symmetric point and use the symmetry principle. But, in this case, I am not sure what these circles would map into. So, we have $(w,1,-1,y)=(z,1,-1,x)$.... and I am not quite sure what I could use for $x$ and $y$. Also, maybe I can't even use $f(1)=1$ and $f(-1)=-1$.... I was only going to use that based on the solution given in that problem. My initial thoughts were to let $a$ be the center of the resulting circles, and then we could use $f(0)=a$ and $f(\frac{1}{4})=a$, but I would still need to find the "$x$" and "$y$" mentioned above.
Any thoughts would be much appreciated!

Answer 1

If $a$ and $b$ are symmetric with respect to both circles $C_1: |z|=1$ and $C_2: |z-\frac{1}{4}|=\frac{1}{4}$ then the Möbius transformation $f$ defined by $$ f(z) = (z, \infty , a, b) = \frac{z-a}{z-b} $$ has the desired properties: $0$ and $\infty$ are symmetric with respect to both $f(C_1)$ and $f(C_2)$ so that both images are circles centered at the origin.

(One could choose $f(z) = (z, c, a, b)$ for arbitrary $c \ne a, b$, that would only multiply $f$ by a constant factor.)

It remains to determine $a$ and $b$. The symmetry of $C_1$ and $C_2$ with respect to the real axis suggest to try it with real numbers: Then

• $a, b$ are symmetric wrt to $C_1$ if $ab=1$, and
• $a, b$ are symmetric wrt to $C_2$ if $(a-1/4)(b-1/4) = 1/4^2$.

It follows that $ab=1$ and $a+b=4$, i.e. $a, b$ are the solution of the quadratic equation $$ z^2 - 4z + 1 = 0 \iff z = 2-\sqrt 3 \, \vee \, z = 2+\sqrt 3 \, . $$ Without loss of generality we can choose $a=2-\sqrt3$ and $b = 2 + \sqrt 3$. Then $$ f(z) = \frac{z-a}{z-b} = \frac{z-(2-\sqrt 3)}{z-(2 + \sqrt 3)} $$ maps $C_1$ and $C_2$ to concentric circles, and the ratio of the radii is $$ \left | \frac{f(-1)}{f(0)}\right| = \frac{(a+1)b}{(b+1)a} = b = 2 + \sqrt 3 \, . $$