# If $z_{1,2,3}\in S(a,R)$ then $z^*={R\over \bar{z}-\bar{a}}+a$

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.

• I think you want $R^2$ in the numerator rather than $R$. – user10354138 Jun 6 at 19:39
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 $$i \, \frac {\overline z - \overline a - R} {\overline z - \overline a + R} = -i \, \frac {z^* - a - R} {z^* - a + R}, \\ z^* = \frac {R^2} {\overline z - \overline a} + a.$$