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 $z_1^2+z_2^2+z_3^2=$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let r be the circum radius of the equilateral triangle and w 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
$O^{{}^{\prime }}A,O^{{}^{\prime }}B,O^{{}^{\prime }}C$ respectively

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

$O{B}^{{}^{\prime }}=z_2-z_0=r{e}^{(\theta +\frac{2\pi }{3})}=rw{e}^{i\theta }$
$O{C}^{{}^{\prime }}=z_3-z_0=r{e}^{(\theta +\frac{4\pi }{3})}=r{w}^2{e}^{i\theta }$
$\therefore z_1=z_0+r{e}^{i\theta },z_2=z_0+rw{e}^{i\theta },z_3=z_0+r{w}^2{e}^{i\theta }$

Squaring and adding
$z_1^2+z_2^2+z_3^2=3z_0^2+2(1+w+w^2)z_{0}r{e}^{ie}+(1+w^2+w^4)r^2{e}^{i2\theta }$
$=3z_0^2,$ Since $1+w+w^2=0=1+w^2+w^4$

Note: Students should rembember this question as a formula.