Question

If $z_1,z_2,z_3$ are complex numbers such that $\frac{2}{z_1}=\frac{1}{z_2}+\frac{1}{z_3}$.

Then prove that the points represented by $z_1,z_2,z_3$ lie on a circle passing through the origin.

Answer 1

$\frac{2}{z_1}=\frac{1}{z_2}+\frac{1}{z_3}$
or $\frac{1}{z_1}-\frac{1}{z_2}=\frac{1}{z_3}-\frac{1}{z_1}$
or $\frac{z_2-z_1}{z_1{z}_2}=\frac{z_1-z_3}{z_3{z}_1}$
or $\frac{z_2-z_1}{z_3-z_1}=\frac{z_2}{z_3}$
$⟹arg\left(\frac{z_2-z_1}{z_3-z_1}\right)=arg(-\frac{z_2}{z_3})$
$arg\left(\frac{z_2-z_1}{z_3-z_1}\right)=\pi +arg\left(\frac{z_2}{z_3}\right)$
or
$arg\left(\frac{z_2-z_1}{z_3-z_1}\right)=\pi -arg\left(\frac{z_2}{z_3}\right)$
or
$⟹\alpha =\pi -\beta$
or $\alpha +\beta =\pi$
Hence, the given points are concyclic.
Ans: 1