Find the possible values of |A + B + C | $ |A |= |B | = |C | = 1 $ ,where A B and C are complex nos
$$ \frac{A^2}{BC}+ \  \frac{B^2}{ \ {CA}} \ +\  \frac{C^2}{ \ {AB}} + 1 = 0$$
Find the possible values of $ |A + B + C |$
Tried substituting cos(theta) + i sin(theta)
 A: $$A=\cos a+i\sin a, B=\cos b+i\sin b, C=\cos c+i\sin c, a, b, c \in [0,2\pi).$$
From the  given condition results:
$$\cos (2a-b-c)+\cos (2b-c-a)+\cos (2c-a-b)+1=0   $$ 
$$\sin (2a-b-c)+\sin (2b-c-a)+\sin (2c-a-b)=0     $$
or $$\cos x+\cos y+\cos z+1=0     $$ 
$$\sin x+\sin y+\sin z=0          $$ where
$$2a-b-c=x, 2b-c-a=y, 2c-a-b=z$$ with $$x+y+z=0$$
$$\cos \frac{x}{2}\cdot\cos \frac{y}{2}\cdot\cos \frac{z}{2}=0  $$
$$\sin \frac{x}{2}\cdot\sin \frac{y}{2}\cdot\sin \frac{z}{2}=0  $$
For $\cos \frac{x}{2}=0 $ and $\sin \frac{y}{2}=0$ 
result $x=(2k+1)\pi, y=2l\pi$
and 
$2a-b-c=(2k+1)\pi, 2b-c-a=2l\pi.$
Find $a=c+\frac{(4k+2l+2)\pi}{3}, b=c+ \frac{(2k+4l+1)\pi}{3}.$
$$A+B+C=\cos (c+\frac{(4k+2l+2)\pi}{3})+\cos (c+\frac{(2k+4l+1)\pi}{3})+\cos c+$$
$$+i[\sin (c+\frac{(4k+2l+2)\pi}{3})+\sin (c+\frac{(2k+4l+1)\pi}{3})+\sin c]=$$
$$= \cos c-2\sin (c+(k+l)\pi)\cdot\cos((k-l)\frac{\pi}{3}+\frac{\pi}{6}+$$
$$+i[\sin c+2\cos (c+(k+l)\pi)\cdot\cos((k-l)\frac{\pi}{3}+\frac{\pi}{6})]$$
$$=\cos c-2(-1)^{k+l}\sin c\cdot\cos((k-l)\frac{\pi}{3}+\frac{\pi}{6}+$$
$$+i[\sin c+2(-1)^{k+l}\cos c\cdot\cos((k-l)\frac{\pi}{3}+\frac{\pi}{6})]$$
$$|A+B+C|=1+4\cos^2 ((k-l)\frac{\pi}{3}+\frac{\pi}{6}) .$$
For $k-l =6p$ or $k-l =6p+4$ find $$|A+B+C|=1.$$
For $k-l =6p+1$ or $k-l =6p+2$ or $k-l =6p+3$ or $ k-l =6p+5$ find $$|A+B+C|=2.$$
