Encuentra los valores posibles de | A + B + C |

Question

$ |A |= |B | = |C | = 1 $, donde AB y C son complejos

ps

Encuentra los valores posibles de$$ \frac{A^2}{BC}+ \ \frac{B^2}{ \ {CA}} \ +\ \frac{C^2}{ \ {AB}} + 1 = 0$

Intenté sustituir cos (theta) + i sin (theta)

Answer 1

$$A=\cos a+i\sin a, B=\cos b+i\sin b, C=\cos c+i\sin c, a, b, c \in [0,2\pi).$ $ De los resultados de la condición dada:$$\cos (2a-b-c)+\cos (2b-c-a)+\cos (2c-a-b)+1=0 $ $$$\sin (2a-b-c)+\sin (2b-c-a)+\sin (2c-a-b)=0 $ $ o$$\cos x+\cos y+\cos z+1=0 $ $$$\sin x+\sin y+\sin z=0 $ $ donde$$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 $ $

Para$\cos \frac{x}{2}=0 $ y$\sin \frac{y}{2}=0$ result$x=(2k+1)\pi, y=2l\pi$

y

$2a-b-c=(2k+1)\pi, 2b-c-a=2l\pi.$

Encontrar $a=c+\frac{(4k+2l+2)\pi}{3}, b=c+ \frac{(2k+4l+1)\pi}{3}.$

$$A+B+C=\cos (c+\frac{(4k+2l+2)\pi}{3})+\cos (c+\frac{(2k+4l+1)\pi}{3})+\cos c+$ $$$+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})]$ $$$|A+B+C|=1+4\cos^2 ((k-l)\frac{\pi}{3}+\frac{\pi}{6}) .$ $ Para$k-l =6p$ o$k-l =6p+4$ find$$|A+B+C|=1.$ $ Para$k-l =6p+1$ o$k-l =6p+2$ o$k-l =6p+3$ o$ k-l =6p+5$ find$$|A+B+C|=2.$ $