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Roots of unity are the solutions of the complex polynomial $t^{n}-1=0$ they have the following form $E_{n}=\{e^{\frac{2\pi ik}{n}}:k\in\mathbb{Z}\}=\{e^{\frac{2\pi ik}{n}}:k=1,...,n-1\}$. From the properties of the $e$-function we know that $|e^{\frac{2\pi ik}{n}}|=1$ for all roots of unity, hence they lie on the unit circle. They are eventually used to construct regular n-gon's.

Now, everywhere I read about them, it's stated that they are evenly spaced around the unit circle. However I'd like to know how to prove this.

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    $\begingroup$ The distance between two complex numbers $z_1$ and $z_2$ is $|z_1-z_2|=\sqrt{(z_1-z_2)(\bar z_1-\bar z_2)}$. Try plugging in two consecutive roots of unity, $z_1=\exp(2\pi ik/n)$ and $z_2=\exp(2\pi i(k+1)/n)=z_1\exp(2\pi i/n)$, and see if the result is independent of $k$. $\endgroup$ – Rahul Feb 13 at 7:12
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    $\begingroup$ It's the $n$ in the denominator of the power that controls the spacing around the circle. Since it never changes, the spacing must be equal $\endgroup$ – postmortes Feb 13 at 7:12
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    $\begingroup$ If you would post this as an answer, I would gladly accept it :) @rahul $\endgroup$ – Christian Singer Feb 13 at 7:14
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    $\begingroup$ Please go ahead and fill in the details, post it yourself, and accept it :) $\endgroup$ – Rahul Feb 13 at 7:15
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    $\begingroup$ See this website: mathonline.wikidot.com/nth-roots-of-unity $\endgroup$ – BadAtGeometry Feb 13 at 7:35
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As pointed out, the roots of

$t^n - 1 = 0 \tag 1$

are the $n$ complex numbers

$\omega^j = e^{2\pi i j / n} = (e^{2\pi i / n})^j, \; 0 \le j \le n - 1. \tag 2$

If we use the Euler identity on the $\omega^j$ we find

$\omega^j = e^{2\pi i j / n} = \cos \dfrac{2\pi j}{n} + i \sin \dfrac{2\pi j}{n}; \tag 3$

it is easy to see from $(3)$ that the ray emanating from the origin and passing through $\omega^j$ makes an angle $2\pi j / n$ with the positive $x$-axis; thus the angle between consecutive roots of unity $\omega^j$ and $\omega^{j + 1}$ is precisely $2\pi /n,$ no matter what the value of $j$; it is the same for any two consecutive $n$-th roots of unity, so the arc subtended by the angle 'twixt two consecutive $\omega^j$ always is of length $2 \pi / n$; they are evenly space around the unit circle.

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  • $\begingroup$ @J. W. Tanner: you my friend are edit like the devil-but so far I've accepted all of them. Cheers! $\endgroup$ – Robert Lewis Feb 13 at 7:44
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    $\begingroup$ Thanks for this nice proof :) $\endgroup$ – Christian Singer Feb 13 at 8:24
  • $\begingroup$ @ChristianSinger: you are most welcome, my friend; and I thank you for the "acceptance"! $\endgroup$ – Robert Lewis Feb 13 at 8:31
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    $\begingroup$ @RobertLewis: Thanks for accepting my edits. They don't call me "eagle eye" for nothing. I actually like your posts $\endgroup$ – J. W. Tanner Feb 13 at 15:50
  • $\begingroup$ @J.W.Tanner: OK, eagle eye, keep up the good work, and thanks for the kind words. You are on of the only "editors" around here whose work I generally trust. Cheers! $\endgroup$ – Robert Lewis Feb 13 at 16:57

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