# Why are roots of unity evenly spaced?

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.

• 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 $k$. – Rahul Feb 13 at 7:12
• 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 – postmortes Feb 13 at 7:12
• If you would post this as an answer, I would gladly accept it :) @rahul – Christian Singer Feb 13 at 7:14
• Please go ahead and fill in the details, post it yourself, and accept it :) – Rahul Feb 13 at 7:15
• See this website: mathonline.wikidot.com/nth-roots-of-unity – BadAtGeometry Feb 13 at 7:35

## 1 Answer

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.

• @J. W. Tanner: you my friend are edit like the devil-but so far I've accepted all of them. Cheers! – Robert Lewis Feb 13 at 7:44
• Thanks for this nice proof :) – Christian Singer Feb 13 at 8:24
• @ChristianSinger: you are most welcome, my friend; and I thank you for the "acceptance"! – Robert Lewis Feb 13 at 8:31
• @RobertLewis: Thanks for accepting my edits. They don't call me "eagle eye" for nothing. I actually like your posts – J. W. Tanner Feb 13 at 15:50
• @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! – Robert Lewis Feb 13 at 16:57