Let
$ \gamma = ${$z \in \mathbb{C} : |z-a| = R$}. Two complex numbers $z_1,z_2$ are said to be symmetric with respect to $\gamma$ iff
$$ (z_1-a)\overline{(z_2-a)} = R^2. $$
I am trying to prove that if $\forall \alpha,\beta \in \gamma$,
$$ \Bigg | \frac{z_1-\alpha}{z_2 – \alpha} \Bigg | = \Bigg |\frac{z_1-\beta}{z_2 – \beta} \Bigg | \qquad \qquad (1)$$
then $z_1 $ and $z_2$ are symmetric with respect to $\gamma$. I was able to prove the converse using the preliminary Möbius transformation
$$ T(z) = \frac{R^2}{z-a} + \overline{a}$$
and properties of the cross-ratio, but, in trying to apply these techniques, I was not able to obtain any substantial progress. I need some help proving this result.
I found the problem in one of my problem sheets in a course in complex analysis mainly based in Silverman's and Lang's textbooks on the subject.
Addendum: Considering a comment that was erased, if one considers the transformation
$$T(z) = \frac{z-\alpha}{z-\beta}$$
where $\alpha,\beta \in \gamma$, $\alpha \neq \beta$, we have, by hypothesis, that if $w_1 = T(z_1)$ and $w_2 = T(z_2)$,
$$ |w_1| = |w_2|$$
thus, if $w = (w_1 + w_2)/2$, $w_1$ and $w_2$ are symmetric with respect to $L = \{ wt : t \in \mathbb{R}\}$, which is a line that "passes through" $0$ and $\infty$, thus, by symmetry preserving of Möbius transformations, $z_1$ and $z_2$ are symmetric with respect to some line or circle which contains $\alpha$ and $\beta$. In addition, $\alpha $ and $\beta$ where chosen arbitrarily from $\alpha$, thus, for any $\alpha, \beta, \delta \in \gamma$, mutually distinct, $z_1$ and $z_2$ are symmetric with respect to a line or circle which contains $\alpha$ and $\beta$ and also with respect to a line or circle which passes through $\alpha$ and $\delta$, however, transitivity of these properties, i.e. deducing that there exists a circle which contains $\alpha,\beta$ and $\gamma$ such that $z_1$ and $z_2$ are symmetric with respect to, seems distant to prove.
Best Answer
Condition $(1)$ means that the cross-ratio $$ (z_1, z_2, \alpha, \beta) = \frac{z_1-\alpha}{z_1-\beta}\cdot\frac{z_2-\beta}{z_2-\alpha} $$ has absolute value $1$ for all $\alpha, \beta$ on the circle $\gamma$. Now let $T$ be a Möbius transformation which maps $\gamma$ to the unit circle with $T(z_1) = 0$. The cross-ratio is invariant under Möbius transformations, so that $w_2 = T(z_2)$ satisfies $$ |(0, w_2, a, b)| = 1 $$ for all $a, b$ on the unit circle. If $w_2 \ne \infty$ then this means that $$ 1 = \left|\frac{0-a}{0-b}\cdot\frac{w_2-b}{w_2-a}\right| = \left|\frac{w_2-b}{w_2-a}\right| $$ or $|w_2-b| = |w_2-a|$ for all $a, b$ on the unit circle. By choosing $(a, b) = (-1, 1)$ and then $(a, b) = (-i, i)$ we can conclude that $w_2$ lies on both the imaginary and on the real axis, to that $w_2 = 0$ and consequently, $z_1 = z_2$.
Otherwise $w_1 = 0$ and $w_2 = \infty$ are symmetric with respect to the unit circle. This symmetry is preserved under the Möbius transformation $T^{-1}$, so that $z_1$ and $ z_2$ are symmetric with respect to $\gamma$.
Conversely, if $z_1, z_2$ are symmetric with respect to $\gamma$ and $T$ maps $\gamma$ to the unit circle with $T(z_1) = 0$ then $T(z_2) = \infty$ and $$ |(0, \infty, a, b)| = \left|\frac{a}{b}\right| = 1 $$ for all $a, b$ on the unit circle, which implies that $$ |(z_1, z_2, \alpha, \beta)| = 1 $$ for all $\alpha, \beta$ on the circle $\gamma$.
This proves that $(1)$ is satisfied for all $\alpha, \beta$ on the circle $\gamma$ if and only if $z_1 = z_2$ or $z_1, z_2$ are symmetric with respect to $\gamma$.