Trigonometric equation with complex numbers Let $x$, $y$, and $z$ be real numbers such that $\cos x+\cos y+\cos z=\sin x+\sin y+\sin z=0$.
Prove that $\cos 2x+\cos 2y+\cos 2z=\sin 2x+\sin 2y+\sin 2z=0$.
Starting with the given equation, I got that $i\sin x+i\sin y+i\sin z=0$. 
Adding this to the other part of the given equation, I then got $\cos x+\cos y+\cos z+i\sin x+i\sin y+i\sin z=0$, which can also be written as $(\cos x+i\sin x)+(\cos y+i\sin y)+(\cos z+i\sin z)=0$.
Here, I let $a=e^{ix}$, $b=e^{iy}$, and $c=e^{iz}$, which, after substituting in to the above equation gives $a+b+c=0$. 
What we want to prove can be written as $a^2+b^2+c^2$, but I am not quite sure how to find this result from what I have.
Some help would be very appreciated. Thanks!

In the same way you have: $(\cos x-i \sin x)+(\cos y-i\sin y)+(\cos z-i \sin z)=0$, it's equal: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, so $ab+ac+bc=0$ and $0=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2=0$
