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

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