Complex numbers algebra problem involving cyclic summation 
Let $a_1$, $a_2$, $a_3\in \mathbb{C}$ and $|a_1|=|a_2|=|a_3|=1$.

If $\sum\frac{a_1^{2}}{a_2 a_3}=-1$, find $|a_1 + a_2 + a_3|$
What I have done till now:
First, I tried to attack the required sum directly. 
Let $\alpha=|a_1 + a_2 + a_3|$ , then squaring both sides we get ,
$$\alpha^{2}=(a_1 + a_2 + a_3)\left(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)$$ since  $|a_1|=|a_2|=|a_3|=1$ and $|z|^{2}= z\overline{z}$, but it did not yield much as i perceived.
Then in the given sum, $\sum\frac{a_1^{2}}{a_2 a_3}=-1$, i tried taking LCM on left side and on solving I got, $a_1^{3} + a_2^{3} + a_3^{3} = -a_1 a_2 a_3$. On manipulations, we get 
$$(a_1 + a_2 + a_3)(a_1^{2} + a_2^{2} + a_3^{2} - a_1 a_2 - a_2 a_3 - a_3 a_1) = 2a_1 a_2 a_3.$$
Here I am facing a dead end. I even tried taking conjugate of $\sum\frac{a_1^{2}}{a_2 a_3}=-1$, and add the 2 equations but it does not seem to be helping much .
Please help me with this problem.
 A: Note that each term in the sum has length $1$. If we add $1$ as a fourth complex number to this sum, we get $0$. In this way, we get a (possibly degenerate) quadrilateral with four sides of equal length, producing a rhombus. In particular this means that pairs of these terms (including $1$) must be negatives of each other. In other words, out of the numbers $\frac{a_1^2}{a_2 a_3}, \frac{a_2^2}{a_1 a_3}, \frac{a_3^2}{a_1 a_2}$, two must sum to $0$, and the other must be $-1$.
Without loss of generality, assume $\frac{a_1^2}{a_2 a_3} = -1 \implies a_1^2 = -a_2a_3$. Then,
$$0 = \frac{a_2^2}{a_1 a_3} + \frac{a_3^2}{a_1 a_2} = a_1(a_2^3 + a_3^3) = a_2^3 + a_3^3 = (a_2 + a_3)(a_2^2 - a_2a_3 + a_3^2).$$
Suppose $a_2 + a_3 = 0$. Then $a_1^2 = a_2^2 \implies a_1 = \pm a_2$. So, $a_1 = -a_2$ or $a_1 = -a_3$, so in either case, $|a_1 + a_2 + a_3| = 1$.
Otherwise, we have $a_2^2 - a_2a_3 + a_3^2 = 0$. Note that
$$(a_2 - a_3)^2 = a_2^2 - a_2 a_3 + a_3^2 - a_2 a_3 = a_1^2.$$
Thus $a_1 - a_2 + a_3 = 0$ or $a_1 + a_2 - a_3 = 0$. In the former case, $a_1 + a_2 + a_3 = 2a_2$, and hence is of length $2$. Similarly, in the latter case, the length is still $2$.
So, in conclusion, the only possible values of $|a_1 + a_2 + a_3|$ are $1$ or $2$.
Let's finish by proving sharpness. If $a_1 = 1$, $a_2 = 1$, and $a_3 = -1$, then the cyclic sum comes to $-1$, and $|a_1 + a_2 + a_3| = 1$. On the other hand, let $a_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$, $a_2 = \overline{a_1} = a_1^{-1}$, and $a_3 = 1$. Then,
$$\frac{a_1^2}{a_2 a_3} + \frac{a_2^2}{a_1 a_3} + \frac{a_3^2}{a_1 a_2} = \frac{a_1^2}{a_2} + \frac{a_2^2}{a_1} + \frac{1}{1} = a_1^3 + a_2^3 + 1 = -1 + -1 + 1 = -1.$$
In this case, $|a_1 + a_2 + a_3| = 2$.
A: We are  given three complex numbers such that 
$|a_1| = |a_2| = |a_3| =1$ that also satisfy
$$\frac{a_1^2}{a_2 a_3} + \frac{a_2^2}{a_1 a_3} + \frac{a_3^2}{a_1 a_2} = -1. \tag{1} $$
Clearing denominators we get
$$ a_1^3 + a_2^3 + a_3^3 = -a_1a_2a_3. \tag{2} $$
Let $$ b_0:=a_1a_2a_3, \, b_1:=a_1^3,\, b_2:=a_2^3,\, b_3:=a_3^3. \tag{3} $$
Now equation $(2)$ can be written as $$ b_0 + b_1 + b_2 + b_3 = 0 \tag{4} $$
where $\,b_0, b_1, b_2, b_3\,$ are the vertices of a (perhaps degenerate)
rectangle. This is because, for example, let $\,M:=(b_0+b_1)/2\,$ be the midpoint of the  chord line segment $\,b_0b_1.\,$ This chord is perpendicular
to the radius passing through $\,M\,$ from the origin, and determines the chord. Now by equation $(4)$ we must have $\, -M = (b_2 + b_3)/2\,$ and replacing $\,M\,$ by $\,-M\,$ negates the pair placing $\,b_2b_3\,$ on the opposite side of the unit circle. If $\,M=0\,$ the two pairs coincide and 
are the vertices of a degenerate rectangle.
Thus there is a pairing of $\,b_0,b_1,b_2,b_3\,$ so that the numbers of
each pair sum to zero. Without loss of generality we renumber $\,b_1,b_2,b_3\,$ so that $\,b_2 = -b_0\,$ and $\,b_3 = -b_1.\,$ By
definition of $\,b_0\,$
$$ b_0^3 = b_1b_2b_3 = b_0b_1^2 \tag{5} $$ which implies $\, b_0^2=b_1^2\,$
and thus $\, b_1^2 = b_2^2 = b_3^2 =: w.\,$  Therefore
$$ \{a_1,a_2,a_3\} \subset \{z\mid z^6=w\}. \tag{6} $$
Without loss of generality, because equation $(2)$ is homogeneous, we may
assume that $\,a_3=1\,$ and that the other two numbers are sixth roots of
unity. A calculation of all possibilities shows that
$$ a_1+a_2+a_3 \in \{1,-1,2,1+\sqrt{-3},1-\sqrt{-3}\}. \tag{7} $$
Finally, $\,|a_1+a_2+a_3|\,$ is either $1$ or $2$.
The $\,-1,1\,$ are both degenerate cases where two of the numbers coincide and the third is their negative. The other cases are when the three number
points on the unit circle are $\,60^\circ\,$ apart from the middle point.
In all of the five cases, the corresponding $\,\{b_0,b_1,b_2,b_3\}\,$ points
are the vertices of a degenerate rectangle.
