# Complex numbers and geometry - Four complex numbers lying on a circle

I'm stuck on a problem of What is Mathematics?, by Courant and Robbins. The formulation is as follows:

Prove that if for four complex numbers $z_1$, $z_2$, $z_3$ and $z_4$, the angles of $\frac{z_3-z_1}{z_3-z_2}$ and $\frac{z_4-z_1}{z_4-z_2}$ are the same, then the four numbers lie on a circle or on a straight line, and conversely.

In a previous exercise I have interpreted the angle of $\frac{z_1-z_2}{z_1-z_3}$ as the (oriented) angle between the vectors $z_1-z_3$ and $z_1-z_2$, so I imagine that I should use some kind of geometrical argument, but I can't find anything useful (probably my geometry knowledge is deficient). I've tried a lot of different algebraic manipulations to show that the four points lie on a circle, but they eventually become extremely cumbersome and I think that the problem can't be that hard...

Any ideas concerning some simple geometric or algebraic argument to show the validity of the proposition?

• The key to the circular part is the Inscribed Angle Theorem (IAT). If $z_1$, $z_2$, $z_3$ aren't collinear, then they're on a (proper) circle. The equality of the two quotients implies (and is implied by) the fact that $\angle z_1 z_4 z_2 = \angle z_1 z_3 z_2$; the IAT says that this fact is equivalent to having $z_4$ live on that circle. (The inscribed angles at $z_3$ and $z_4$ "look at" the same arc, $\stackrel{\frown}{z_1z_2}$.)
– Blue
Dec 27, 2014 at 13:54
– MvG
Dec 27, 2014 at 19:13

Let $z_1$, $z_2$ and $z_3$ be non-collinear.
Now sketch a circle passing through $z_1$, $z_2$ and $z_3$.
Take $\overline{z_1 z_2}$ as a chord.
Now looking for angle $\widehat{z_1 z_2 z_3} = \theta$, the chord subtends the same angle at every point on the circle (and not any other point). Now for angle $\widehat{z_1 z_4 z_2} = \theta$, it has to lie on the circle: hence, first part is proved.
Second case, $z_1$, $z_2$ and $z_3$ are collinear and thus $\theta$ is $0$ or $\pi$.
Thus angle $\widehat{z_1 z_4 z_2}$ is either zero or $\pi$ implying them to be collinear and thus all four to be collinear.