Question

Let the complex numbers $z_1,z_2 \,$ and $\, z_3$ be the vertices of an equilateral triangle. Let $z_0$ be the circumcentre of the triangle. Then, prove that $z_1^2+z_2^2+z_3^2=3z_0^2$

Answer 1

Since, $z_1,z_2,z_3$ are the vertices of an equilateral triangle.
$\therefore $ Circumcentre $(z_0)=$ Centroid $\bigg(\frac{z_1+z_2+z_3}{3} \bigg).....(i)$
Also, for equilateral triangle
$z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1 ....(ii)$
On squaring Eq. (i), we get
$9z_0^2=z_1^2+z_2^2+z_3^2+2(z_1+z_2+z_2z_3+z_3z_1)$
$\Rightarrow 9z_0^2=z_1^2+z_3^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1) $ [from Eq. (ii)]
$\Rightarrow 3z_0^2=z_1^2+z_2^2+z_3^2$