Question

Let $z_1,z_2,z_3$ be complex numbers such that $z_1^2+z_2^2+z_3^2=z_1{z}_2+z_2{z}_3+z_3{z}_1$ and $|z_1+z_2+z_3|=21.$ Given that $|z_1-z_2|=2\sqrt{3},|z_1|=3\sqrt{3},$ then the value of $|z_2{|}^2+|z_3{|}^2$ is

Answer 1

Since $z_1\ne z_2\ne z_3$
$\therefore \sum z_1^2=\sum z_1{z}_2\Rightarrow z_1,z_2,z_3$ are the vertices of equilateral triangle.
Now, $|z_1-z_2|=|z_2-z_3|=|z_3-z_1|=2\sqrt{3}$
Given, $|z_1+z_2+z_3|=21$
$\Rightarrow (z_1+z_2+z_3)(\overline{z_1}+\overline{z_2}+\overline{z_3})=441$
$\Rightarrow |z_1{|}^2+|z_2{|}^2+|z_3{|}^2+(z_1\overline{z_2}+z_2\overline{z_1})+(z_2\overline{z_3}+z_3\overline{z_2})+(z_3\overline{z_1}+z_1\overline{z_3})=441$

Let $(z_1\overline{z_2}+z_2\overline{z_1})+(z_2\overline{z_3}+z_3\overline{z_2})+(z_3\overline{z_1}+z_1\overline{z_3})=x$
Then, $|z_2{|}^2+|z_3{|}^2=441-|z_1{|}^2-x$
$\Rightarrow |z_2{|}^2+|z_3{|}^2=441-27-x$
$\Rightarrow |z_2{|}^2+|z_3{|}^2=414-x\cdots \left(1\right)$

Now, $|z_1-z_2{|}^2=12\cdots \left(2\right)$
$|z_2-z_3{|}^2=12\cdots \left(3\right)$
$|z_3-z_1{|}^2=12\cdots \left(4\right)$
On adding $\left(2\right),\left(3\right)$ and $\left(4\right),$ we get
$x=2(|z_2{|}^2+|z_3{|}^2)+18\cdots \left(5\right)$

$\therefore$ From $\left(1\right)$ and $\left(5\right),$ we get
$3(|z_2{|}^2+|z_3{|}^2)=396$
$\Rightarrow |z_2{|}^2+|z_3{|}^2=132$