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?

We will use some facts such as $|a|=1$ implies $a\bar{a}=1$. So $\frac{1}{a}=\bar{a}$. Also let $w=a+b+c$, then $|w|^2=(a+b+c)(\bar{a}+\bar{b}+\bar{c})$.
\begin{align*}
\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}&=-1\\
\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}&=-1\\
a^3+b^3+c^3+abc & =0\\
a^3+b^3+c^3-3abc & =-4abc\\
(a+b+c)(a^2+b^2+c^2-ab-bc-ca)&=-4abc\\
(a+b+c)((a+b+c)^2-3ab-3bc-3ca)&=-4abc.
\end{align*}
 we get
\begin{align*}
w^3-3w(ab+bc+ca)&=-4abc\\
w^3-3wabc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)&=-4abc\\
w^3&=abc\left[3w\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-4\right]\\
w^3&=abc\left[3w\left(\bar{a}+\bar{b}+\bar{c}\right)-4\right]\\
w^3&=abc\left[3|w|^2-4\right]\\
|w^3|&=\left|3|w|^2-4\right|\\
\tag{1}
\label{eq1}
|w|^3&=\left|3|w|^2-4\right|\\
\end{align*}


*

*Case 1: If $3|w|^2-4 \geq 0$, then \eqref{eq1} becomes $|w|^3-3|w|^2+4=0$ and this implies $|w|=2$.

*Case 2: If $3|w|^2-4 < 0$, then \eqref{eq1} becomes $|w|^3+3|w|^2-4=0$ and this implies $|w|=1$.


Thus $|w| \in \{1,2\}$.
