If $\Gamma$ is a circle through points $z_2,z_3,z_4$ then $z^*$ and $z$ in $\Bbb C_\infty$ are said to be symmetric if

$(z^*,z_2,z_3,z_4)$= Conjugate of $(z,z_2,z_3,z_4)$ (I couldn't get an overline on this bracket)

If $\Gamma$ is a straight line then $z^*$ and $z$ should be symmetric w.r.t. $\Gamma$ if the line through $z^*$ and $z$ is perpendicular to $\Gamma$ and $z^*$ and $z$ are at the same distance from $\Gamma$ but on the opposite sides of $\Gamma$. Am I correct ?

If I am, then if both $z^*$ and $z$ lie on $\Gamma$, then that would mean that they are not symmetric ?

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    $\begingroup$ Two points of $\Gamma$ are symmetric with respect to $\Gamma$ if and only if they coincide. $\endgroup$ – Did Sep 4 '14 at 11:51
  • $\begingroup$ @Did I don't understand how can they be symmetric if they coincide? $\endgroup$ – johny Sep 4 '14 at 12:11
  • $\begingroup$ As usual, if and only the symmetry with respect to $\Gamma$ (defined on the whole space, especially also on $\Gamma$ itself) sends one on the other. $\endgroup$ – Did Sep 4 '14 at 12:13

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