Question

Let the complex number $z_1,z_{{}_2},z_3$ be the vertices of an equilateral triangle. Let z$z_0$ be the circumcentre of the triangle , then proove that ${z_1}^2+{z_2}^2+{z_3}^2=3{z_0}^2$

Answer 1

we have,

In an equilateral triangle $z_1,z_2,z_3$ be the vertices.

And $z_0$ be the circum centre of the triangle.

Then we know that,

$z_0=\frac{z_1+z_2+z_3}{3}$

$z_1+z_2+z_3=3z_0$

Squaring both side and we get,

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

${z_1}^2+{z_2}^2+{z_3}^2+2z_1{z}_2+2z_2{z}_3+2z_1{z}_3=9{z_0}^2$

${z_1}^2+{z_2}^2+{z_3}^2+2(z_1{z}_2+z_2{z}_3+z_1{z}_3)=9{z_0}^2......\left(1\right)$

But this triangle is equilateral then,

${z_1}^2+{z_2}^2+{z_3}^2=2(z_1{z}_2+z_2{z}_3+z_1{z}_3)$

Now, by equation (1) and we get,

${z_1}^2+{z_2}^2+{z_3}^2+2({z_1}^2+{z_2}^2+{z_3}^2)=9{z_0}^2$

$3({z_1}^2+{z_2}^2+{z_3}^2)=9{z_0}^2$

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

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

Hence, this is the answer.