Question

Let $z_1,z_2$ and $z_3$ are three complex numbers such that $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1$ and $\frac{z_1^2}{z_2{z}_3}+\frac{z_2^2}{z_3{z}_1}+\frac{z_3^2}{z_1{z}_2}+1=0$ then find the sum of all possible values of $\left|z_1+z_2+z_3\right|$.

Answer 1

Answer: 3

$\Theta \frac{z_1^2}{z_2{z}_3}+\frac{z_2^2}{z_3{z}_1}+\frac{z_3^2}{z_1{z}_2}+1=0$
$\Rightarrow z_1^3+z_2^3+z_3^3+z_1{z}_2{z}_3=0$
$\Rightarrow \left(z_1+z_2+z_3\right)\left({\left(z_1+z_2+z_3\right)}^2-3\left(z_1{z}_2+z_2{z}_3+z_3{z}_1\right)\right)=-4z_1{z}_2{z}_3$
$\Rightarrow {\left(z_1+z_2+z_3\right)}^3=z_1{z}_2{z}_3\left(3\left(z_1+z_2+z_3\right)\left(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right)-4\right)$
$=z_1{z}_2{z}_3\left(3\left(z_1+z_2+z_3\right)\left({\overline{z}}_1+{\overline{z}}_2+{\overline{z}}_3\right)-4\right)$
$\therefore {\left|z_1+z_2+z_3\right|}^2=\left|z_1{z}_2{z}_3\right||3|z_1+z_2+{z_3|}^2-4\mid$
Let$\mid z_1+z_2+z_3\mid =x$
$\therefore x^3=\left|3x^2-4\right|$
$\Rightarrow x=1\text{ or }2\Rightarrow 1+2=3$