trigonometric system In order to show that $ e^{ix}+e^{iy}+e^{iz}=0 \Longrightarrow e^{2ix}+e^{2iy}+e^{2iz}=0 $, I want to prove that $ \cos x+\cos y+\cos z=0 $ and $ \sin x+\sin y+\sin z=0 \Longrightarrow \cos 2x+\cos 2y+\cos 2z=0$ and $ \sin 2x+\sin 2y+\sin 2z=0 $
$ \cos 2x=2\cos^2 x-1=2(\cos y+\cos z)^2-1 $
$ \cos 2x+\cos 2y+\cos 2z=2(\cos^2x+\cos^2y+\cos^2z+$ 
$(\cos^2x+cos^2y+\cos^2z+2(\cos x\cos y+\cos y\cos z+\cos x\cos z)))-3 $
$ \cos 2x+\cos 2y+\cos 2z=2(\cos^2x+\cos^2y+\cos^2z)-3 =3-2(\sin^2x+\sin^2y+\sin^2z)$ ...
Any idea?

You can prove it with pure complex number manipulations itself (and if needed a "pure trigonometric" proof can be read off from that).
Hint:
If $z_1 + z_2 + z_3 = 0$ then $\overline{z_1} + \overline{z_2} + \overline{z_3} = 0$ and if $|z| = 1$, then $\overline{z} = \frac{1}{z}$.
Now try squaring something and use the above two facts.
Since OP seems to ignore this, will forget about giving further hints. For the sake of completeness:
Spoiler

 If $z_1 = e^{ix}$, $z_2 = e^{iy}$ and $z_3 = e^{iz}$, then we have $z_1 + z_2 + z_3 = 0$. We also have $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} = 0$ which gives $z_1 z_2 + z_2 z_3 + z_3 z_1 = 0$, multiplying by $z_1z_2z_3$. Squaring the first gives $z_1^2 + z_2^2 + z_3^2 + 2 (z_1 z_2 + z_2 z_3 + z_3 z_1) = 0$, which implies $z_1^2 + z_2^2 + z_3^2 = 0$ which is what you wanted to prove.

