What can be said about complex numbers $z_1, z_2, z_3$ if $\frac{z_1 - z_3}{z_2 - z_3}$ is real?

Question

$z_1, z_2, z_3 \in \mathbb{C}$ $$\frac{z_1 - z_3}{z_2 - z_3} \in \mathbb{R}$$

The only idea I'm coming up with is that either $ \operatorname{Im}(z_1) = \operatorname{Im}(z_2) = \operatorname{Im}(z_3)$ or $ \operatorname{Re}(z_1) = \operatorname{Re}(z_2) = \operatorname{Re}(z_3)$. If I were to do the division properly, i.e. express all complex numbers in the $( x+iy )$ format and multiply both the numerator and denominator by the conjugate of $z_2 - z_3$, I'd end up with 6 variables, which seems very complicated. Do you have any ideas as to how to solve this? thanks!

Answer 1

It means the points are collinear, since the paths from $z_3$ to the other $z_i$ are either parallel or antiparallel. (We also require $z_2\ne z_3$.)

In particular the phases of $z_1-z_3,\,z_2-z_3$ either match if the ratio is positive, or differ by $\pi$ if it is negative; or, if it $0$, $z_1=z_3$ so collinearity is trivial. Conversely, in the event of collinearity the ratio is trivially real.

Answer 2

Geometrically, it means the affixes of $z_1,z_2$ and $z_3$ are collinear.

Algebraically, let $\,t$ be the (real) value of this ratio. We have $$z_1-z_3=t(z_2-z_3)\iff z_1=tz_2+(1-t)z_3,$$ i.e. $z_1$ is a barycentre of $z_2$ and $z_3$, with weights $t$ and $1-t$.