Problem
Let $C$ be a circle or a line belonging to $\overline{\mathbb C}$ and let $z_2,z_3,z_4$. Two points $z$ and $z^*$ are said to be symmetric with respecto to $C$ if $\overline{(z,z_2,z_3,z_4)}=(z^*,z_2,z_3,z_4)$.
$(i)$ Prove that the previous definition doesn't depend on the chosen points $z_2,z_3,z_4 \in C$ but of $C$.
$(ii)$ Prove that for each $z \in \overline{\mathbb C}$ there is a unique symmetric point $z^*$ with respect to $C$. The function that assigns to each $z$ its correspondent $z^*$ with respect to $C$ is called symmetry with respect to $C$. Show that for each Möbius transformation $T$ which maps $\overline{\mathbb R}$ to $C$, the function $$T \circ \overline{T^{-1}}:\overline{\mathbb C} \to \overline{\mathbb C}$$ is the symmetry with respect to $C$.
My attempt
$(i)$ Let $C$ be a circle centered at $c$ of radius $R$ Using invariance of cross ratio under Möbius transformations, and using that $z_i-c=R$ for $i=2,3,4$ and $z\overline{z}=|z|^2$ we get $$\overline{(z,z_2,z_3,z_4)}=\overline{(z-c,z_2-c,z_3-c,z_4-c)}=(\overline{z}-\overline{c},\overline{z_2-c},\overline{z_3-c},\overline{z_4-c})=(\overline{z}-\overline{c},\dfrac{R^2}{z_2-c},\dfrac{R^2}{z_3-c},\dfrac{R^2}{z_4-c})=(\dfrac{R^2}{\overline{z}-\overline{c}},z_2-c,z_3-c,z_4-c)=(\dfrac{R^2}{\overline{z}-\overline{c}}+c,z_2,z_3,z_4)$$
So if $C$ is a circle, from this equation one deduces the dependence only on $C$.
$(ii)$ If $C$ is a circle, from the formula $z^*=\dfrac{R^2}{\overline{z}-\overline{c}}+c$ it follows the uniqueness of $z^*$.
I need help to show $(i)$ and uniqueness of $C$ if $C$ is a line. I also don't know what to do to show that $T \circ \overline{T^{-1}}:\overline{\mathbb C} \to \overline{\mathbb C}$ is the symmetry with respect to $C$, I would appreciate any suggestions.
Best Answer
Let $C$ be any circle and $R$ be the real line
[I'll use $z^*_C$ to denote the symmetric point of $z$ with respect to the $C$, and $z^* = z^*_R$ to denote the complex conjugate of $z$.]
Let $f(z) = (z,a;b,c)$ where $a,b,c$ are distinct points in $C$
Let $g(z) = (z,d;e,f)$ where $d,e,f$ are distinct points in $C$
Then $f$ is an invertible Mobius transformation that sends $(a,b,c)$ to $(1,0,\infty)$ and hence $C$ to $R$
Similarly for $g$
Let $h = g f^{-1}$
Then $h$ is an invertible Mobius transformation from $R$ to itself
Thus $h(z) = \frac{pz+q}{rz+s}$ for some $p,q,r,s \in \mathbb{R}$ because:
$\frac{p}{r} = h(\infty) \in R$ and hence WLOG $p,r \in \mathbb{R}$
$h(z) = 0$ for some $z \in R$ and hence $s \in \mathbb{R}$
$\frac{q}{s} = h(0) \in R$ and hence $q \in \mathbb{R}$
Thus $h(z^*) = \frac{pz^*+q}{rz^*+s} = \frac{p^*z^*+q^*}{r^*z^*+s^*} = (\frac{pz+q}{rz+s})^* = h(z)^*$
For any $z,z^*_C$ such that $f(z^*_C) = f(z)^*$:
$g(z^*_C) = g(f^{-1}(f(z^*_C))) = h(f(z)^*) = h(f(z))^* = g(z)^*$
Therefore the definition is independent of the three distinct points in $C$
For any Mobius transformation $T$ that maps $R$ to $C$:
Let $U = T^{-1} f^{-1}$ [You could use $U = f T$ and it would work as well.]
Then as before $U(z^*) = U(z)^*$
$T(T^{-1}(z)^*) = T(T^{-1}(f^{-1}(f(z)))^*) = T(U(f(z))^*) = T(U(f(z)^*)) = f^{-1}(f(z)^*) = z^*_C$
The uniqueness of the symmetric points was already evident from the invertibility of $f$