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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?

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You might start by noting that without loss of generality you can assume $z = 0$. –  Robert Israel Aug 22 '11 at 16:13
    
If $ z=0 $, $ cosx+cosy=-1, sinx=sin(-y), x=-y (mod2\pi) $ or $ x=\pi+y (mod2\pi) $, if $ x=-y $, $ cosx=-1/2, x=2\pi/3 $ or $ -2\pi/3 (mod2\pi), y=-2\pi/3 $ or $ 2\pi/3 (mod2\pi), sin(4\pi/3)+sin(-4\pi/3)=0 $, $ cos(4\pi/3)+cos(-4\pi/3)=-1 $ ... but why can we write $ z=0 $ ? Is it due to the fact that there are two equations and three unknown quantities ? –  Chon Aug 22 '11 at 17:34
1  
@Plane Chon-Ju: You can take $z=0$ because you can multiply each side of the equation by $e^{-iz}$. Then let $u=x-z$, $v=y-z$. –  André Nicolas Aug 22 '11 at 18:08

1 Answer 1

up vote 6 down vote accepted

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.

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+1 for use of >! to hide the spoiler. Neat! –  Sasha Aug 22 '11 at 23:33

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