Question

Let $z_1,z_2,\&z_3$ be the vertices and $z_0$ be the circumcentre of an equilateral triangle. then $z_1^2+z_2^2+z_3^2$ =

• A. $z_0^2$
• B. $-z_0^2$
• C. $3z_0^2$
• D. $-3z_0^2$

Answer 1

The correct option is C $3z_0^2$

Let r be the circum radius of the equilateral triangle and $\omega$ the cube root of unity.

Let ABC be the equilateral triangle with $z_1,z_2$ and $z_3$ as its vertices A, B and C respectively with circumcentre $O^{\prime }\left(z_0\right)$. The vectors $O^{\prime }A,O^{\prime }B,O^{\prime }C$ are equal and parallel to $O^{\prime }A,O^{\prime }B,O^{\prime }C$ respectively.

Then the vectors $\overline{O{A}^{\prime }}$ = $z_1-z_0$ = $r{e}^{i\theta }$

$\overline{O{B}^{\prime }}$ = $z_2-z_0$ = $r{e}^{(\theta +\frac{2\pi }{3})}$ = $r\omega e^{i\theta }$

$\overline{O{C}^{\prime }}$ = $z_3-z_0$ = $r{e}^{i(\theta +\frac{4\pi }{3})}$ = $r{\omega }^2{e}^{i\theta }$

$\therefore z_1$ = $z_0+r{e}^{i\theta },z_2$ = $z_0+r\omega e^{i\theta }$, $z_3$ = $z_0+r{\omega }^2{e}^{i\theta }$

Squaring and adding

$z_1^2+z_2^2+z_3^2=3z_0^2+2(1+\omega +{\omega }^2)z_{0}r{e}^{i\theta }+(1+{\omega }^2+{\omega }^4)r^2{e}^{i2\theta }$

$=3z_0^2$, Since $1+\omega +{\omega }^2$ = 0 = $1+{\omega }^2+{\omega }^4$