<s>[INST] 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.
 [/INST] 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, $|a_1 + a_2 + a_3| = 2$.
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