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Recall:

  • $z^*\in\mathbb{C}$ is symetric to $z\in\mathbb{C}$ in a relation to a generalized circle $C$ if $\overline{(z_1,z_2,z_3,z)}=(z_1,z_2,z_3,z^*)$ for some $z_1,z_2,z_3\in C$.
  • $(z_1,z_2,z_3,z)\in\mathbb{C}$ is defined to be the point $T(z)$ where $T$ is a Mobius transformation such that $T(z_1)=0,T(z_2)=1,T(z_3)=\infty$

The question:

Suppose $C=S(a,R)$. (I assume that $S(a,R)$ is a circle of radii $R$ and center $a$. Show that $$ z^*={R\over \bar{z}-\bar{a}}+a$$

My attampt:

We know that $$T(z)={z-z_1\over z-z_3}:{z_2-z_1\over z_2-z_3}$$ Let $$A:=z_2-z_3, B:=z_1(z_2-z_3)\\C:=z_2-z_1,D:=z_2-z_1$$ Thus, $$T(z)={Az-B\over Cz-D}\Rightarrow \\\overline{T(z)}={\bar{A}\bar{z}-\bar{B}\over \bar{C}\bar{z}-\bar{D}} = {\bar{A}\bar{z}-\bar{B}+B-B\over \bar{C}\bar{z}-\bar{D}+D-D}\\ ={ ({\bar{A}\bar{z}-\bar{B}+B \over A})\cdot A-B \over ({\bar{C}\bar{z}-\bar{D}+D \over C })\cdot C -D} $$ So I brought $\overline {T(z)}$ to the form of $T$ but I don't know how to proceed.

Thanks on advence.

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    $\begingroup$ I think you want $R^2$ in the numerator rather than $R$. $\endgroup$ – user10354138 Jun 6 at 19:39
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You need to use the condition that $z_i$ lie on the circle. Take $z_1 = a + R, \, z_2 = a + i R, \, z_3 = a - R$. Then $\overline {T(z)} = T(z^*)$ gives $$i \, \frac {\overline z - \overline a - R} {\overline z - \overline a + R} = -i \, \frac {z^* - a - R} {z^* - a + R}, \\ z^* = \frac {R^2} {\overline z - \overline a} + a.$$

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