On a Geometric Proof (Ahlfors) that the Cross ratio is real if and only if four points lie on a circle or straight line

Question

The cross-ratio $(z_1,z_2,z_3,z_4)$ is real if and only if the four points lie on a circle or on a straight line.

I know this question has been asked numerous times on MSE, but I have a specific question with respect to how Ahlfors proves this on page 79 of his book. He says,

"This is evident by elementary geometry, for we obtain $$\text{arg}(z_1,z_2,z_3,z_4)=\text{arg}\frac{z_1-z_3}{z_1-z_4}-\text{arg}\frac{z_2-z_3}{z_2-z_4}$$ and if the points lie on a circle this difference of angles is either $0$ or $\pm\pi$"

Could someone elaborate a bit on this for me? I see that he's splitting up the cross-ratio using arguments, and I know that straight lines and circles are the same in $\mathbb{R}\cup\infty$, but I am not seeing how the geometry proves both directions of the statement. I would appreciate it very much! Thank you.

Answer 1

In the diagram above, by the inscribed angle theorem, the angles at $z_1,z_2$ are equal because they subtend the same chord $z_3z_4$. If $z_1,z_2$ are on opposite sides of the chord the sum of the subtended angles is $\pi$.

Recalling that $\text{arg}\frac{w}{z}$ is the angle between $w$ and $z$, in the first case (angles equal), $$\text{arg}\frac{z_1-z_3}{z_1-z_4}=\alpha=\text{arg}\frac{z_2-z_3}{z_2-z_4}.$$ and from there we see that $\text{arg}(z_1,z_2,z_3,z_4)=0 \implies (z_1,z_2,z_3,z_4)\in\mathbb{R}$.

The argument is similar if $z_1,z_2$ are on opposite sides of the chord.

To go in the other direction we observe that $(z_1,z_2,z_3,z_4)\in\mathbb{R}\implies \text{arg}(z_1,z_2,z_3,z_4)=0 \text{ or }\pi,$ and therefore the angles at $z_1,z_2$ subtended by the segment $z_3z_4$ are either equal or supplementary. In either case we can conclude that the points are either collinear or concyclic.

Bonus mnemonic. The cross ratio formula can be difficult to remember, whether it is for the complex plane, or for collinear points in projective geometry. This geometric interpretation gives a good mnemonic. In the diagram there are two paths to get from $z_3$ to $z_4$ via the other points: $z_3\to z_1\to z_4$, and $z_3\to z_2\to z_4$. For each path take the ratios of the path segments, and then the ratio of these ratios: $$ \frac{z_3\to z_1}{z_1\to z_4}:\frac{z_3\to z_2}{z_2\to z_4}$$

Rewriting $w\to z$ as $w-z$ we get (*) $$ \frac{z_3-z_1}{z_1-z_4}:\frac{z_3-z_2}{z_2-z_4},$$

which is equivalent to Ahlfors' definition

$$ \frac{z_1-z_3}{z_1-z_4}:\frac{z_2-z_3}{z_2-z_4}.$$

(*) Technically, instead of $w-z$ we should have rewritten $w\to z$ in 'vector' style as $z-w$ but $z-w=-(w-z)$ and all the minus signs cancel. This way we get a cross ratio formula that's easy to 'read' as two paths from $z_3$ to $z_4$.