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

Answer 1

The correct option is C $3z_0^2$

Theorem - the triangle whose vertices are

the points repressed by the

compels numbers $z_1,z_2,z_3$ on the

argent plane is equilateral if

$\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0$

i.e. iff $z_1^2+z_2^2+z_3^2=z_1{z}_2+z_2{z}_3+z_3{z}_1...\left(A\right)$

Circumcentre and centroid of equilateral

triangle coincides.

Thus, $z_0=\frac{1}{2}(z_1+z_2+z_3)$

$\Rightarrow 3z_0=z_1+z_2+z_3$

$\Rightarrow (3z_0{)}^2=(z_1+z_2+z_3{)}^2$

$=z_1^2+z_2^2+z_3^2+2(z_1{z}_2+z_2{z}_3+z_3{z}_1)...\left(B\right)$

As the triangle is equilateral

$z_1{z}_2+z_2{z}_3+z_3{z}_1=z_1^2+z_2^2+z_3^2...\left(2\right)$

${97}_0^2=3(z_1^2+z_2^2+z_3^2)$ [from (1) and (2)]

$3z_0^2=z_1^2+z_2^2+z_3^2$

So, option (c) is correct.