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.
$...$
can throw off the spacing, which defeats part of the purpose of mathematical type-setting. It's better (and easier) to do$2+2=4$
than$2$ $+$ $2$ $=$ $4$
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