I need to prove that 4 points of the complex plane lie on a same line or a same circle if and only the following is right for corresponding complex numbers:

$$\frac{z_1-z_3}{z_2-z_3} \div \frac{z_1-z_4}{z_2-z_4} \text{ is real.}$$

$\frac{z_1-z_3}{z_2-z_3} \div \frac{z_1-z_4}{z_2-z_4} \in \mathbb{R} \Leftrightarrow \frac{z_1-z_3}{z_2-z_3}$ and $\frac{z_1-z_4}{z_2-z_4}$ both real or both imaginary. If they are both real, then we have $z_1-z_3 = r_1(z_2-z_3)$ and $z_1-z_4 = r_2(z_2-z_4), r_1,r_2 \in \mathbb{R}$. So vectors $\vec{Z_1}-\vec{Z_3},\vec{Z_2}-\vec{Z_3}$ and $\vec{Z_1}-\vec{Z_4},\vec{Z_2}-\vec{Z_4}$ are collinear.

And what if $\dfrac{z_1-z_3}{z_2-z_3}$ and $\dfrac{z_1-z_4}{z_2-z_4}$ are both imaginary?

  • $\begingroup$ It is not correct that $a/b\in\mathbb R \Longleftrightarrow{}$ $a$ and $b$ are either both real or both imaginary. It is true that $a/b\in\mathbb R \Longleftarrow{}$ $a$ and $b$. However, suppose $a = 2+6i$ and $b=1+3i$. Then $a/b$ is real even though neither $a$ nor $b$ is real and neither $a$ nor $b$ is imaginary. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 3 '15 at 12:29
  • $\begingroup$ $\uparrow$ I didn't finish a sentence: I meant $a/b\in\mathbb R \Longleftarrow{}$ $a$ and $b$ are both real or $a$ and $b$ are both imaginary. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 3 '15 at 12:46

I would try this:

  • Show that for any three distinct complex numbers $z_1,z_2,z_3$ there is a linear fractional transformation $z\mapsto\dfrac{az+b}{cz+d}$ for which $\dfrac a b \ne \dfrac c d$ that maps $z_1,z_2,z_3$ to distinct real numbers.

  • Show that if $z_4$ is on the same circle or the same line as $z_1,z_2,z_3$ then $z_4$ is also mapped to a real number. I.e. show that the image of a line or a circle under a linear fractional transformation is a line or a circle.

  • Show that if $w_i$ is the image of $z_i$ under the linear fractional transformation for $i=1,2,3,4$, then $$\frac{w_1-w_3}{w_2-w_3} \div \frac{w_1-w_4}{w_2-w_4}= \frac{z_1-z_3}{z_2-z_3} \div \frac{z_1-z_4}{z_2-z_4}.$$ I.e. linear fractional transformations do not change cross-ratios.

Maybe the second item is where most of the work needs to be done.

  • $\begingroup$ The "cross ratio" comes up in projective geometry. A very tiny nit to pick with Michael Hardy ; Say " $ ad \ne bc $ ", not "$ a/b \ne c/d$ " because $ b $ or $ d $ can be $ 0$. $\endgroup$ – DanielWainfleet Sep 3 '15 at 17:08

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