<s>[INST] If $a,b,c$ are three complex numbers Find possible values of $\lvert a+b+c \rvert$ Given three complex numbers $a,b,c$ such that $\lvert a \rvert=\lvert b \rvert=\lvert c \rvert=1$ and $$\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1$$
Find which of the following are possible values of $\lvert a+b+c \rvert$
A)0
B)2
C)1.5
D)3
My try:
I assumed $a=e^{ix}$,$b=e^{iy}$, $c=e^{iz}$
Then we have
$$\cos (2x-y-z)+\cos (2y-x-z)+\cos (2z-x-y)=-1$$
$$\sin(2x-y-z)+\sin (2y-x-z)+\sin (2z-x-y)=0$$
Squaring and adding we get
$$3+2(\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=1$$
So
$$\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=-1$$
any clue here?
 [/INST] The four complex numbers$\frac{a^2}{bc},\frac{b^2}{ac},\frac{c^2}{ab},1$ all have length $1$ and sum to zero. Added as vectors,'head to tail', in the Argand diagram they form a closed shape and must consist of  two pairs of equal but  opposite vectors. 
Without loss of generality we therefore have $\frac{a^2}{bc}=-\frac{b^2}{ac},\frac{c^2}{ab}=-1$ and then $a^3=-b^3,ab=-c^2$. Write $c$ as $wa$, then $b=-w^2a$,  where $w$ must be a 6th root of unity.
Then $|a+b+c|=|1+w-w^2|$ is $2$ if $w$ is a primitive 3rd or 6th root and is $1$ if $w$ is $1$ or $-1.$ The answer is B.
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