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$z_1$, $z_2$, and $z_3$ are 3 complex numbers. Prove that if they represent the vertices of an equilateral triangle then $z_1 + \omega z_2 + \omega^2 z_3 = 0$ where $\omega$ is a 3rd root of unity. Any help would be thoroughly appreciated.

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    $\begingroup$ Does not hold for $(z_1,z_2,z_3)=(1,\omega^2,\omega)$, you need to fix the orientation to be the same as $(1,\omega,\omega^2)$. And $\omega$ must be a primitive 3rd root of unity, i.e. $\omega\neq1$. $\endgroup$ – ccorn Sep 27 '14 at 5:47
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    $\begingroup$ Wait, I get it: Then there exists a 3rd root of unity, $\omega$, such that ... This is true, as $\omega$ can be chosen according to the orientation of $(z_1,z_2,z_3)$. $\endgroup$ – ccorn Sep 27 '14 at 6:32
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    $\begingroup$ it's a rather subtle case of "read the question very carefully"! however, as per mathematical convention, the association of $\omega$ with a particular value of $\frac{\sqrt{3}i-1}2$ is very close, so if your interpretation is correct the question could perhaps be judged a little unfair. $\endgroup$ – David Holden Sep 27 '14 at 7:11
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    $\begingroup$ If any of the answers below were useful to you, then you should upvote (if you can) all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. $\endgroup$ – gebruiker Nov 10 '17 at 15:48
  • $\begingroup$ This one has been open for a while, wonder if the user is still active. $\endgroup$ – Wesley Strik Sep 6 '18 at 21:28
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Hint: Consider the expression $a = z_1 + \omega z_2 + \omega^2 z_3$. It is invariant under translation ($(z_1,z_2,z_3) \mapsto (z_1+z,z_2+z,z_3+z)$) since $1+\omega+\omega^2=0$, and it is homogeneous, that is $a(zz_1,zz_2,zz_3) = za(z_1,z_2,z_3)$. Since all equilateral triangles can be transformed to one another using translation and rotation/scaling (the operation corresponding to multiplying all points by some complex number), it is enough to prove the result for some fixed equilateral triangle. Choose one and do the math.

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Yuval's answer cannot be improved on for elegance, but a slightly messier solution may be worth a footnote

(taking note of @ccorn's shrewd insight we'll here explicitly number the vertices anticlockwise, i.e. co-$\omega$-wise). it is easy to see geometrically that the equilateral symmetry can be expressed in terms of invariance of the figure under the action of $C_3$ represented as rotations about the centroid of the triangle.

since multiplication by $\omega$ effects an anti-clockwise rotation through $\frac{2\pi}3$ for the triangle to be equilateral requires: $$ \left(z_1 - \frac{z_1+z_2+z_3}3 \right)\omega = z_2 - \frac{z_1+z_2+z_3}3 $$ i.e. $$ (2\omega+1)z_1-(2+\omega)z_2+(1-\omega)z_3 = 0 $$ and the result follows on multiplication by $\frac{1+2\omega^2}3$

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