Question

Let the complex numbers z₁,z₂ and z₃ be the vertices of an equilateral triangle. Let z₀ be the circumcentre of the triangle, then $z_1^2+z_2^2+z_3^2$=

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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 = $1+{\omega }^2+{\omega }^4$