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I'm struggling to see for a method to start this question.
It looks like a question related with mobius transforms.
We have studied about determining the mobius transform when points from the domain are given. For example, getting the cross ratio.

But here I don't see any points from the domain and also a method to determine the center when the mobius transform is given.

Help would be appreciated

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  • $\begingroup$ Hint: $r$ can be anything greater than half the distance from $0$ to $2+i$. $\endgroup$ – Robert Israel Apr 15 at 0:23
  • $\begingroup$ Can you please explain how should I proceed afterwards $\endgroup$ – Charith Jeewantha Apr 15 at 1:38
  • $\begingroup$ Given $r$, $z$ could be anything in the intersection of the disks of radius $r$ around $0$, $1$ and $2+i$. $\endgroup$ – Robert Israel Apr 15 at 2:28
  • $\begingroup$ On the other hand, you could first choose any $z$, and then let $r$ be greater than the maximum of $|z|$, $|z-1|$ and $|z-(2+i)|$. $\endgroup$ – Robert Israel Apr 15 at 2:35
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    $\begingroup$ OK, so you want $z$ to be equidistant from $0$, $1$ and $2+i$. Construct the perpendicular bisectors of the line segments $[0,1]$ and $[1, 2+i]$, and see where they intersect. $\endgroup$ – Robert Israel Apr 15 at 4:07
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With $z = x +iy$ you have the three equations

  1. $\lvert z - 0 \rvert = r$. i.e. $x^2 + y^2 = r^2$

  2. $\lvert z - 1 \rvert = r$, i.e. $(x - 1)^2 + y^2 = x^2 - 2x + 1 + y^2 = r^2$

  3. $\lvert z - (2+i) \rvert = r$, i.e. $(x - 2)^2 + (y - 1)^2 = (x - 2)^2 + y^2 - 2y + 1 = r^2$

Subtracting 2.from 1. you get $2x - 1 = 0$, hence $x = \frac{1}{2}$. Thus $y^2 = r^2 - \frac{1}{4}$. Inserting this in 3. yields $\frac{9}{4} + r^2 - \frac{1}{4} - 2y +1 = r^2$ which means $y = \frac{3}{2}$.

Then you get $\frac{1}{4} + \frac{9}{4} = r^2$, i.e. $r = \sqrt{\frac{5}{2}}$.

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