Four complex numbers $z_1,z_2,z_3,z_4$ lie on a generalized circle if and only if they have a real cross ratio $[z_1,z_2,z_3,z_4]\in\mathbb{R}$

Question

Let $[z_1,z_2,z_3,z_4]$ denote the cross ratio of the complex numbers $z_1,z_2,z_3,z_4\in \mathbb{C}$. Show that the distinct points $z_1,z_2,z_3,z_4\in\widehat{\mathbb{C}}$ lie on a generalized circle if and only if $[z_1,z_2,z_3,z_4]\in\mathbb{R}$.

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I saw this statement claimed in this answer, and wondered how one would go about showing it. I know of some facts we could work with:

• First, the definition: A generalized circle in $\mathbb{C}$ is either a circle or a line in $\mathbb{C}$.
• I know that a Mobius transformation maps generalized circles to generalized circles.
• I know that if $f$ is a Mobius transformation and $z_j'=f(z_j)$, $j=1,2,3,4$, then $[z_1',z_2',z_3',z_4']=[z_1,z_2,z_3,z_4]$.

I have posted a potential proof as an answer below. Please let me know if you think it's complete. Thank you!

Answer 1

We will show (1) if $z_1,z_2,z_3,z_4$ lie on a generalized circle, then $[z_1,z_2,z_3,z_4]\in\mathbb R$, and (2) if $[z_1,z_2,z_3,z_4]\in\mathbb R$, then $z_1,z_2,z_3,z_4$ lie on a generalized circle.

1. First, suppose $z_1,z_2,z_3,z_4$ lie on a generalized circle. We know that, given three real numbers $x_1,x_2,x_3 \in \mathbb{R}$, there exists a Mobius transformation \begin{equation*} F=F_{x_1,x_2,x_3}^{-1} \circ F_{z_1,z_2,z_3} \end{equation*} such that $F(z_i)=x_i$, $i=1,2,3$. Also, since Mobius transformations map generalized circles to generalized circles, we know that $z_4$ is also mapped to some $x_4 \in \mathbb{R}$. Therefore, since $x_1,x_2,x_3,x_4 \in \mathbb{R}$, then their cross ratio $[x_1,x_2,x_3,x_4]=\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \in \mathbb{R}$. Therefore, since the cross ratio is invariant under Mobius transformation, we have \begin{align*} [z_1,z_2,z_3,z_4] &=[F(z_1),F(z_2),F(z_3),F(z_4)] \\ &=[x_1,x_2,x_3,x_4] \\ &=\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \\ &\in \mathbb{R} \end{align*}
2. Next, suppose $[z_1,z_2,z_3,z_4] \in \mathbb{R}$. Let \begin{equation*} F=F_{x_1,x_2,x_3}^{-1} \circ F_{z_1,z_2,z_3} \end{equation*} be the Mobius transformation which sends $F(z_i)=x_i$, $i=1,2,3$, and let $z_4'=F(z_4)$. Since the cross ratio is invariant under Mobius transformation, we know that \begin{equation*} [z_1,z_2,z_3,z_4]=[x_1,x_2,x_3,z_4'] \end{equation*} Therefore the cross ratio $[x_1,x_2,x_3,z_4']$ is also a real number, which implies that $z_4'$ is a real number as well. Now consider $F^{-1}$, which is also a Mobius transformation. Since $x_1,x_2,x_3,z_4'$ lie on a generalized circle (the real line), and since a Mobius transformation maps generalized circles to generalized circles, then $z_1,z_2,z_3,z_4 \in F^{-1}(\mathbb{R})$ belong to a generalized circle.

Answer 2

$\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}}$ Lemma: If $|z|=|w|$, then $$ \arg(z-w)=\frac{\arg(z)+\arg(w)}2\pm\frac\pi2\tag1 $$ Proof: Let $z=re^{ia}$ and $w=re^{ib}$. $$ \begin{align} re^{ia}-re^{ib} &=\color{#C00}{2}r\,\frac{e^{i(a-b)/2}-e^{i(b-a)/2}}{\color{#C00}{2}\color{#090}{i}}\,\color{#090}{e^{i\pi/2}}e^{i(a+b)/2}\tag{2a}\\ &=2r\sin\left(\frac{a-b}2\right)\,e^{i(a+b+\pi)/2}\tag{2b} \end{align} $$ Since $\sin\left(\frac{a-b}2\right)$ might be positive or negative, the argument might be $\frac{a+b+\pi}2$ or $\frac{a+b-\pi}2$.

$\large\square$

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Invariance of the Cross Ratio

A cross ratio is unchanged by the translation (represented by $b$), rotation (represented by $\arg(a)$), and scaling (represented by $|a|$) in the map $z\mapsto az+b$. That is, $$ \begin{align} \frac{((az_1+b)-(az_3+b))\ ((az_2+b)-(az_4+b))}{((az_2+b)-(az_3+b))\ \ ((az_1+b)-(az_4+b))} &=\frac{a(z_1-z_3)\ a(z_2-z_4)}{a(z_2-z_3)\ a(z_1-z_4)}\tag{3a}\\ &=\frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}\tag{3b} \end{align} $$

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Moving Three Points to the Real Line or the Unit Circle

For three collinear points, translate one of the points to the origin and rotate the others to the real axis.

For three non-collinear points, translate their circumcenter to the origin, then scale so the points are all on the unit circle.

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Mapping a Fourth Point

Define $$ \begin{align} f(\color{#C00}{z}) &=\frac{(z_1-z_3)(z_2-\color{#C00}{z})}{(z_2-z_3)(z_1-\color{#C00}{z})}\tag{4a}\\ &=\frac{z_1-z_3}{z_2-z_3}\left(\frac{z_2-z_1}{z_1-\color{#C00}{z}}+1\right)\tag{4b} \end{align} $$ Then, as long as $z_1,z_2,z_3$ are not coincident, $f$ is bijective on the Riemann Sphere ($\mathbb{C}^\ast$). In fact, $$ \begin{align} f^{-1}(\color{#C00}{w}) &=\frac{\color{#C00}{w}\,z_1(z_2-z_3)-z_2(z_1-z_3)}{\color{#C00}{w}\,(z_2-z_3)-(z_1-z_3)}\tag{5a}\\ &=z_1\frac{\color{#C00}{w}-\frac{z_2}{z_1}\frac{z_1-z_3}{z_2-z_3}}{\color{#C00}{w}-\frac{z_1-z_3}{z_2-z_3}}\tag{5b} \end{align} $$ As mentioned above, any three points can be moved either to the real line or to the unit circle using a map $z\mapsto az+b$. Furthermore, the map $z\mapsto az+b$ preserves the cross ratio.

Therefore, all that needs to be considered is either $\Im(z_1)=\Im(z_2)=\Im(z_3)=0$ (three points on the real axis) or $|z_1|=|z_2|=|z_3|=1$ (three points on the unit circle).

Case $\boldsymbol{1}$: $\boldsymbol{\Im(z_1)=\Im(z_2)=\Im(z_3)=0}$

Since $z_1,z_2,z_3\in\mathbb{R}$, $f:\mathbb{R}^\ast\to\mathbb{R}^\ast$ and $f^{-1}:\mathbb{R}^\ast\to\mathbb{R}^\ast$ (where $\mathbb{R}^\ast=\mathbb{R}\cup\{\infty\}$).

Thus, if three of four points are collinear, then the cross-ratio of the four points is real iff all four points are collinear.

Case $\boldsymbol{2}$: $\boldsymbol{|z_1|=|z_2|=|z_3|=1}$

Let's first prove a Corollary of the Lemma.

Corollary: Suppose $|z_1|=|z_2|=|z_3|$. Then $$ \frac{z_2}{z_1}\frac{z_1-z_3}{z_2-z_3}=\overline{\frac{z_1-z_3}{z_2-z_3}}\tag6 $$ Proof: $$ \begin{align} \arg\left(\frac{z_1-z_3}{z_2-z_3}\right) &=\left(\frac{\arg(z_1)+\arg(z_3)}2\pm\frac\pi2\right)-\left(\frac{\arg(z_2)+\arg(z_3)}2\pm\frac\pi2\right)\tag{7a}\\ &\equiv\frac{\arg(z_1)-\arg(z_2)}2\pmod{\pi}\tag{7b} \end{align} $$ Explanation:
$\text{(7a):}$ apply the Lemma
$\text{(7b):}$ $\pm\frac\pi2-\pm\frac\pi2\equiv0\pmod{\pi}$

Therefore, $$ \begin{align} \arg\left(\frac{z_2}{z_1}\frac{z_1-z_3}{z_2-z_3}\right)+\arg\left(\frac{z_1-z_3}{z_2-z_3}\right) &=\arg\left(\frac{z_2}{z_1}\right)+2\arg\left(\frac{z_1-z_3}{z_2-z_3}\right)\tag{8a}\\[3pt] &\equiv\arg(z_2)-\arg(z_1)\\[3pt] &+2\left(\frac{\arg(z_1)-\arg(z_2)}2\pmod{\pi}\right)\tag{8b}\\[3pt] &\equiv0\pmod{2\pi}\tag{8c} \end{align} $$ Explanation:
$\text{(8a):}$ multiplication is associative
$\text{(8b):}$ $\arg\left(\frac{z_2}{z_1}\right)=\arg(z_2)-\arg(z_1)$
$\phantom{\text{(8b):}}$ apply $(7)$
$\text{(8c):}$ $2(x\pmod{\pi})=2x\pmod{2\pi}$

Since $|z_1|=|z_2|$, we have $$ \left|\frac{z_2}{z_1}\frac{z_1-z_3}{z_2-z_3}\right|=\left|\frac{z_1-z_3}{z_2-z_3}\right|\tag9 $$ $(8)$ and $(9)$ show that $\frac{z_2}{z_1}\frac{z_1-z_3}{z_2-z_3}$ and $\frac{z_1-z_3}{z_2-z_3}$ have the same absolute value and opposite arguments. Thus, they are conjugates.

$\large\square$

Suppose $|z|=1$, then $$ \begin{align} \arg(f(z)) &=\arg\left(\frac{(z_1-z_3)(z_2-z)}{(z_2-z_3)(z_1-z)}\right)\tag{10a}\\ &=\left(\frac{\arg(z_1)+\arg(z_3)}2\pm\frac\pi2+\frac{\arg(z_2)+\arg(z)}2\pm\frac\pi2\right)\\ &-{}\left(\frac{\arg(z_2)+\arg(z_3)}2\pm\frac\pi2+\frac{\arg(z_1)+\arg(z)}2\pm\frac\pi2\right)\tag{10b}\\[3pt] &\equiv0\pmod\pi\tag{10c} \end{align} $$ Explanation:
$\text{(10a):}$ definition of $f$
$\text{(10b):}$ apply the Lemma
$\text{(10c):}$ $\pm\frac\pi2+\pm\frac\pi2\equiv0\pmod{\pi}$

That is, the cross-ratio of $4$ concyclic points is real.

Suppose that $w\in\mathbb{R}$, then $$ \begin{align} \left|f^{-1}(w)\right| &=\left|\,z_1\frac{w-\frac{z_2}{z_1}\frac{z_1-z_3}{z_2-z_3}}{w-\frac{z_1-z_3}{z_2-z_3}}\,\right|\tag{11a}\\ &=\left|\,\frac{\overline{w}-\overline{\frac{z_1-z_3}{z_2-z_3}}}{w-\frac{z_1-z_3}{z_2-z_3}}\,\right|\tag{11b}\\[9pt] &=1\tag{11c} \end{align} $$ Explanation:
$\text{(11a):}$ apply $\text{(5b)}$
$\text{(11b):}$ $|z_1|=1$ and $\overline{w}=w$ since $w\in\mathbb{R}$
$\phantom{\text{(11b):}}$ apply the Corollary
$\text{(11c):}$ $\left|\overline{z}\right|=|z|$

That is, if the cross-ratio of $4$ non-collinear points is real, they are concyclic.

Answer 3

First, show that every generalized circle can be sent by a Mobius transformation to the real line (transitivity). Then, all you have to do is to show that if the cross ratio of $4$ real points is always real.

Also note that this theorem can be shown with elementary geometry using the inscribed angle theorem.

Answer 4

Define $T(z)= [z,z_2,z_3,z_4]$.

$(\Rightarrow)$ Case $1$: $z_i$'s lie on a straight line.

In this case, $\arg(z_i-z_1)= \arg(z_2-z_1) \, \forall i\in \{2,3,4\}.$ It follows that $\arg T(z_1)=2\pi k\implies T(z_1)\in \mathbb R$.

Case $2$: $z_i$'s lie on a circle.

In this case, calculations show that $T(z_1)= \overline{T(z_1)}\implies T(z_1)\in \mathbb R.$

For the other direction, the following theorem would make things straightforward.

Theorem A Mobius transformation maps straight lines and circles onto straight lines and circles.

Remarks:

$(1)$ The theorem doesn't say that a straight line necessary goes to a straight line, similarly for a circle.

$(2)$ Consider the map $f: \mathbb C_\infty\to \mathbb C_\infty:z\mapsto 1/z$. Consider the equation of a generalised circle $a(x^2+y^2)+bx+cy+d=0.$ Writing $z=x+\iota y, f(z)= u+\iota v= \frac{1}{x+\iota y}\implies x=\frac{u}{u^2+v^2}, y=\frac{v}{u^2+v^2}$. Setting these in the equation of the circle: $d(u^2+v^2)+bu+cv+a=0$, which is a generalised circle. So $f$ maps generalised circles to generalise circles.

Pf. of the theorem: A Mobius transformation is a composition of translation, dilation and f. Translation and dilation map generalised circles to generalised circles and $f$ does the same too by Remark $(2)$. Thus, the statement of the theorem follows.

$(\Leftarrow)$ Suppose that $T(z_1)\in \mathbb R$. Clearly, $T(z_2)=1, T(z_3)=0$. Consider the Mobius transformation $S:= T^{-1}$. By the Theorem above, $S$ maps the $x-$ axis onto a generalised circle. It follows that $S(T(z_1)), S(T(z_2)), S(T(z_3)), S(T(z_4))$, which are $z_1,z_2,z_3,z_4$ respectively, lie on a generalised circle.