Question

Prove that the points $z_1,z_2,z_3,z_4$ taken in order are concyclic if and only if $\frac{(z_3-z_1)(z_4-z_2)}{(z_3-z_2)(z_4-z_1)}$ is purely real.

Answer 1

If $ABCD$ be concyclic then angle subtended by $AB$ both at
$C$ and $D$ are same.
$\frac{z_2-z_3}{z_1-z_3}=\frac{BC}{AC}{e}^{i\theta }$
$\frac{z_2-z_4}{z_1-z_4}=\frac{BD}{AD}{e}^{i\theta }$
Dividing, we get
$\left(\frac{z_2-z_3}{z_1-z_3}\right).\left(\frac{z_1-z_4}{z_2-z_4}\right)=\frac{BC.AD}{AC.BD}=$ real
Conversely $\frac{(z_2-z_3)(z_1-z_4)}{(z_1-z_3)(z_2-z_4)}=real=k$ say
then $\frac{z_2-z_3}{z_1-z_3}=k\frac{z_1-z_4}{z_2-z_4}$
$\therefore arg\frac{z_2-z_3}{z_1-z_3}=arg\frac{z_1-z_4}{z_2-z_4}$
$\therefore \angle ACB=\angle ADB$
Hence the points are concyclic.A
Note: If arg $z=\theta$ then arg $kz=\theta$, where $k$ is
real.