Question

Let $z_1,z_2,z_3$ be three complex numbers such that $|z_1|=|z_2|=|z_3|=1$ and $\frac{z_1^2}{z_2{z}_3}+\frac{z_2^2}{z_1{z}_3}+\frac{z_3^2}{z_1{z}_2}=-1.$ Then the possible value(s) of $|z_1+z_2+z_3|$ is/are

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B), (C)

$2$

$\frac{z_1^2}{z_2{z}_3}+\frac{z_2^2}{z_1{z}_3}+\frac{z_3^2}{z_1{z}_2}=-1$
$\Rightarrow z_1^3+z_2^3+z_3^3+z_1{z}_2{z}_3=0$

Now, by using the relation $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ we can find $(z_1+z_2+z_3)$

$\Rightarrow -4z_1{z}_2{z}_3=z_1^3+z_2^3+z_3^3-3z_1{z}_2{z}_3$
$\Rightarrow -4z_1{z}_2{z}_3=(z_1+z_2+z_3)(z_1^2+z_2^2+z_3^2-z_1{z}_2-z_2{z}_3-z_3{z}_1)$

Let $z=z_1+z_2+z_3$
$\Rightarrow z^3-3z(z_1{z}_2+z_2{z}_3+z_3{z}_1)=-4z_1{z}_2{z}_3$
$\Rightarrow z^3=z_1{z}_2{z}_3[3z(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3})-4]$
$\Rightarrow z^3=z_1{z}_2{z}_3\left[3z\right(\overline{z_1}+\overline{z_2}+\overline{z_3})-4]$
$\Rightarrow z^3=z_1{z}_2{z}_3(3|z{|}^2-4)$

Taking the absolute values of both sides, we get
$|z{|}^3=|3|z{|}^2-4|(\because |z_1|=|z_2|=|z_3|=1)$
If $|z|\ge \frac{2}{\sqrt{3}}$, then
$|z{|}^3-3|z{|}^2+4=0$
$\Rightarrow |z|=2$

If $|z|<\frac{2}{\sqrt{3}}$, then
$|z{|}^2+3|z{|}^2-4=0$
$\Rightarrow |z|=1$