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

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!

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  • $\begingroup$ Without loss of generality you can assume $z_3=0$. Maybe that helps with your interpretation. $\endgroup$ – Diger Feb 10 at 20:11
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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.

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  • $\begingroup$ What does 'antiparallel' mean? $\endgroup$ – Pedro Tamaroff Feb 10 at 20:08
  • $\begingroup$ @PedroTamaroff Pointing in opposite directions. Consider what happens if the ratio is positive, or negative. $\endgroup$ – J.G. Feb 10 at 20:09
  • $\begingroup$ Sure. I just had never heard the term before. To make this answer better, perhaps you can provide with a (sketch of) proof of your claim. $\endgroup$ – Pedro Tamaroff Feb 10 at 20:19
  • $\begingroup$ @PedroTamaroff I've added some more detail. $\endgroup$ – J.G. Feb 10 at 20:24
  • $\begingroup$ Thank you very much! If you were required to provide at least some calculations to show how you arrived at the solution, what would you do? Would the polar form prove useful in any way? $\endgroup$ – Minah Feb 10 at 20:36
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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 $1-t$.

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