If $\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then $|z_{1}+z_{2}+z_{3}|$ 
If $z_{1},z_{2},z_{3}$ are three complex number such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$
and $\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then possible values of $|z_{1}+z_{2}+z_{3}|$

$\bf{My\; Try::}$ Given $$\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1\Rightarrow z^3_{1}+z_{2}^3+z_{3}^3=-z_{1}z_{2}z_{3}$$
Now Using $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
So $$\left(z_{1}+z_{2}+z_{3}\right)\left[z^2_{1}+z^2_{2}+z^2_{3}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}\right] = -4z_{1}z_{2}z_{3}$$
So $$(z_{1}+z_{2}+z_{3})\cdot ((z_{1}+z_{2}+z_{3})^2-3(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})) = -4z_{1}z_{2}z_{3}.......(1)$$
Now Let $k=z_{1}+z_{2}+z_{3}.$ Taking Conjugate, we get $\displaystyle k = \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}} = \frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}z_{2}z_{3}}$
So we get $z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=kz_{1}z_{2}z_{3}$
Put into $(1)\;,$ We get $$(z_{1}+z_{2}+z_{3})\cdot \left[(z_{1}+z_{2}+z_{3})^2-3kz_{1}z_{2}z_{3}\right] = -4z_{1}z_{2}z_{3}$$
Now How can I solve after that, Help Required, Thanks
 A: Here is a 3rd solution (inspired by yours, @Christian Blatter, but using different variables). 
Let  $$Z_1=\frac{z^2_{1}}{z_{2}z_{3}}, \ Z_2=\frac{z^2_{2}}{z_{3}z_{1}}, Z_3=\frac{z^2_{3}}{z_{1}z_{2}} \ \ \ (*)$$
Let us remark that all $Z_k \in \mathbb{U}$ (unit circle).
Thus the initial condition is replaced by $Z_1+Z_2+Z_3=-1 \ \ \ (1)$
Another immediate relationship is $Z_1Z_2Z_3=1 \ \ \ (2)$
At last, is we set $w=Z_1Z_2+Z_2Z_3+Z_3Z_1$, we have
$$w=\frac{z_{1}z_{2}}{z^2_{3}}+\frac{z_{3}z_{1}}{z^2_{2}}+\frac{z_{2}z_{3}}{z^2_{1}}$$
Using the rule $u \in\mathbb{U} \Longrightarrow \bar{w}=\dfrac{1}{w}$ :
$\bar{w}=Z_1+Z_2+Z_3$ thus $w=-1 \ \ \ (3) \ $ (by using (1))
Taking into account (1), (2), (3), by Vieta's formulas, $\{Z_1,Z_2,Z_3\}$ is the set of roots of 
$$Z^3+Z^2-Z-1=0$$
i.e., $1$ (multiplicity 1) and $-1$ (multiplicity 2).
One can assume, using definitions (*), that $Z_1=1,Z_2=Z_3=-1$ ; thus
$$\begin{cases}z_1^2=z_2z_3\\ z_2^2=-z_3z_1\\ z_3^2=-z_1z_2\end{cases}  \ \ \ (4)$$
or, equivalently:
$$\dfrac{z_1}{z_2}=\dfrac{z_3}{z_1}=-\dfrac{z_2}{z_3}=:\zeta \ \ \ (5)$$
(i.e., we have given the name $\zeta$ to this common ratio).
Thus, by elementary operations on (5), one finds 
$$\zeta^3=-1 \ \ \  \Longrightarrow \ \ \zeta=-1, e^{i \pi/3}, e^{-i \pi/3} \ \ \ (6)$$
Thus, using (5), the set of solutions of the initial equation is  
$$\begin{cases}z_1=z_1\\ z_2=\zeta^2 z_1\\ z_3=\zeta z_1\end{cases} \ \ \ (7)$$
where $z_1$ is any element of $\mathbb{U}$.
Thus $|z_1+z_2+z_3|=|z_1||1+\zeta+\zeta^2|=2|z_1|=2$ whatever the value of $\zeta$.
A: (You made a mistake when taking conjugates: You have $\bar k=\ldots\ $.)
Denote by $s_1$, $s_2$, $s_3$ the elementary symmetric functions of the $z_i$. It is easy to see that we may assume $s_3=1$. Then the $z_i$ are the solutions of the third degree equation
$$z^3-s_1z^2+s_2z-1=0\ .\tag{1}$$
The solutions of the equation $z^3-\bar s_1z^2+\bar s_2z-1=0$ are then the $\bar z_i$, and the solutions of $1-\bar s_1w+\bar s_2 w^2- w^3=0$, resp.
$$w^3-\bar s_2w^2+\bar s_1 w-1=0\tag{2}$$ are the $\bar z_i^{-1}=z_i$ again. Comparing $(1)$ and $(2)$ we see that necessarily $s_2=\bar s_1$.
Now it is well known that the power sum $p_3:=\sum z_i^3$ can be expressed in terms of the $s_j$ as
$$p_3=s_1^3-3s_1s_2+3s_3\ .$$
The given condition then amounts to
$s_1^3-3s_1\bar s_1+3=-1$, or
$$s_1^3=3|s_1|^2-4\ .$$
Writing $s_1=re^{i\phi}$ with $r\geq0$ we see that necessarily $e^{3i\phi}\in\{1, -1\}$, so that either $r^3-3r^2+4=0$, or $r^3+3r^2-4=0$, which then leads to $r\in\{1,2\}$. 
Both these values can be realized: Letting $z_1=z_2=-1$, $z_3=1$ the given condition is satisfied with $|z_1+z_2+z_3|=1$, and letting
$z_1=1$, $z_2=e^{i\pi/3}$, $z_3=e^{-i\pi/3}$ leads to $|z_1+z_2+z_3|=2$.
