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I'm trying to find $r>0$ such that there is a Mobius transformation between $\left\{z\in\Bbb C:r<|z|<1\right\}$ and the domain bounded by $|z-1/4| = 1/4$ and $|z| =1$. I have no clue what $r$ should be. Any hint for this problem?

one potato two potato
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    You can proceed similarly as in this problem: https://math.stackexchange.com/q/3213413/42969. The idea is to find two points $a, b$ which are symmetric with respect to both circles $|z-1/4| = 1/4$ and $|z|=1$, and map these points to $0, \infty$. – Martin R Oct 14 '22 at 13:00
  • @MartinR Are there such two points? I can't see. The only symmetry is with respect to the real line. – one potato two potato Oct 14 '22 at 13:09
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    Two points $a,b$ are symmetric with respect to the circle with center $c$ and radius $r$ if $(a-c)\overline{(b-c)} = r^2$, see for example https://mathworld.wolfram.com/SymmetricPoints.html. As an example, $a$ and $1/\bar a$ are symmetric with respect to the unit circle – Martin R Oct 14 '22 at 13:13
  • @MartinR Oh, that's clever. So the general philosophy is 'any Mobius transformation preserves symmetric points'. I see there should be a unique $r$. The symmetric points I found are $2\pm\sqrt{3}$. So the map should be of the form $T(z) = c{z-(2-\sqrt{3})\over z-(2+\sqrt{3})}$ for some constant $c$. Since I want the unit circle to be mapped to the unit circle, $|T(1)| = 1$ so $c = 2+\sqrt{3}$. Since the inner circle $|z-1/4| = 1/4$ mapped to an inner circle of an annulus, $|T(0)| = {1\over 2+\sqrt{3}}$ so $r$ should be ${1\over 2+\sqrt{3}}$. Is the basic flow of the argument correct? – one potato two potato Oct 14 '22 at 13:47
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    I did not verify your exact numbers, but yes, that is the “basic flow” :) – Martin R Oct 14 '22 at 13:49
  • See also https://math.stackexchange.com/q/977765/42969. – Martin R Oct 14 '22 at 16:41
  • Also: https://math.stackexchange.com/q/1026127/42969 – Martin R Oct 14 '22 at 16:51
  • @MartinR Hello, I found another post: https://math.stackexchange.com/a/4673/668308 which deals with the same kind of problem but a bit different approach. – one potato two potato Dec 07 '22 at 15:17

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