Franklin's Notes

Cross ratio

The cross ratio of four distinct complex numbers $z_1,z_2,z_3,z_4\in\mathbb C$ is defined as If $z_2,z_3,z_4$ are distinct, then the function is the unique mobius transformation that sends $z_2\mapsto 1$, has a zero at $z_3\mapsto 0$, and has a pole at $z_4\mapsto \infty$.

Proposition 1. The cross-ratio $\rho(z_1,z_2,z_3,z_4)$ is purely real iff the four points $z_1,z_2,z_3,z_4\in\mathbb C$ lie on a common circle or line.

Proof. Let $z_1,z_2,z_3,z_4$ be points on the same circle, and consider the following diagram:

If the labelled arc from $z_1$ to $z_2$ has a measure of $2\gamma$ radians, we have that both of the angles $\angle z_1 z_3 z_2$ and $\angle z_1 z_4 z_2$ have measure $\gamma$, by a theorem from elementary geometry which says that the arc subtended by an angle inscribed in a circle has twice the measure of the inscribed angle. Now, consider the ratios and We have that the argument of the complex number $\rho_1$ is precisely the measure of the angle $\angle z_1 z_3 z_2$, and similarly the argument of the complex number $\rho_2$ is precisely the measure of the angle $\angle z_1 z_4 z_2$. Both of these are equal to $\gamma$, so that $\rho_1$ and $\rho_2$ have the same argument, and therefore $\rho_1/\rho_2$ is real. Now, notice that Hence, for any $4$ points lying on the same circle, the cross-ratio takes a real value. This is, of course, true for $4$ points on the same line as well, since the arguments of $\rho_1$ and $\rho_2$ would both be equal to either $0$ or $\pi$ in this case.

To see why the reverse direction holds, we may notice that if we consider $\rho$ as a function of $z_4$ with $z_1,z_2,z_3$ being fixed, we have that it is a mobius transformation when $z_1,z_2,z_3$ are distinct, making it bijective. Further, every possible real value of $\rho$ is assumed as $z_4$ ranges over the circle or line containing $z_1,z_2,z_3$, meaning that each possible real value of $\rho$ is attained at precisely one complex number $z_4$, and that complex number lies on the circle or line in question. $\blacksquare$




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