Question

If $z_1,z_2,z_3$ are the vertices of equilateral triangle with centroid $z_0$, then $z_1^2+z_2^2+z_3^2$ equals

• A. $z_0^2$
• B. $2z_0^2$
• C. $3z_0^2$
• D. $9z_0^2$

Answer 1

Answer: (C)

$3z_0^2$

If $z_0$ is the centroid of the triangle then $z_0=\frac{z_1+z_2+z_3}{3}$

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

To find $z_1^2+z_2^2+z_3^2$

we have ${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

or $a^2+b^2+c^2={(a+b+c)}^2-2(ab+bc+ca)$

$\therefore z_1^2+z_2^2+z_3^2={(z_1+z_2+z_3)}^2-2(z_1{z}_2+z_2{z}_3+z_3{z}_1)$

$\Rightarrow z_1^2+z_2^2+z_3^2={\left(3z_0\right)}^2-2(z_1{z}_2+z_2{z}_3+z_3{z}_1)$

$\Rightarrow z_1^2+z_2^2+z_3^2=9{z_0}^2-2(z_1{z}_2+z_2{z}_3+z_3{z}_1)$ ..........$\left(1\right)$

In an equilateral triangle all the sides are equal.

$\therefore AB=BC=CA=a$

$\because \angle ABC=\frac{\pi }{3}$

From coni method,

$\frac{z_1-z_2}{z_3-z_2}=\frac{a}{a}{e}^{\frac{i\pi }{3}}$ ....$\left(2\right)$

and $\angle BAC=\frac{\pi }{3}$

From coni method,

$\frac{z_3-z_1}{z_2-z_1}=\frac{a}{a}{e}^{\frac{i\pi }{3}}$ ....$\left(3\right)$

From equations $\left(2\right)$ and $\left(3\right)$, we get

$\frac{z_1-z_2}{z_3-z_2}=\frac{z_3-z_1}{z_2-z_1}$

$\Rightarrow (z_1-z_2)(z_1-z_2)=(z_3-z_2)(z_3-z_1)$

$\Rightarrow z_1^2+z_2^2+z_3^2=z_1{z}_2+z_2{z}_3+z_3{z}_1$ on simplification

Substituting for $z_1{z}_2+z_2{z}_3+z_3{z}_1=z_1^2+z_2^2+z_3^2$ in eqn$\left(1\right)$ we get

$\Rightarrow z_1^2+z_2^2+z_3^2=9{z_0}^2-2(z_1{z}_2+z_2{z}_3+z_3{z}_1)$

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

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

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