Find the possible values of |A + B + C |

Question

$ |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)

Answer 1

$$A=\cos a+i\sin a, B=\cos b+i\sin b, C=\cos c+i\sin c, a, b, c \in [0,2\pi).$$ 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 $$|A+B+C|=2.$$