# In solving $n^{\text{th}}$ roots of unity, for $e ^{2ki\pi}$ , $\;k = 0,\pm1, \pm2, \ldots$ , why must “k” must be an integer?

In solving $n^{\text{th}}$ roots of unity, when using the expression $e ^{2ki\pi}$ , $\;k = 0,\pm1, \pm2, \ldots$ , why must "$k$" must be an integer?

I understand there's a substitution going on, but why must "$k$" be an integer? Wouldn't that limit the range of the angle? I suppose in the original expression $z = e ^ {in\theta}$ where $\theta$ is an arbitrary angle. Why would a substitution using $2k\pi$ where k is an integer be equivalent as $n\theta$? As $n$ is also an integer, wouldn't that suggest that the original $\theta$ must also be a multiple of $2\pi$?

If I just ignore the "why" and keep going on, I can very well understand the process of solving roots of unity, but I am just a bit curious about this "why". I've searched for Wikipedia and textbooks and stuff but they all seem to not have a explanation...

Thanks a lot!

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It is $e^{2k \pi i/n}$. –  André Nicolas Nov 19 '12 at 18:27
"Finding $n^{\text{th}}$ roots of unity" means solving the equation $z^n = 1$. It's clear from taking absolute values that that $|z| = 1$, so write $z = e^{i\theta}$. Now we have $e^{in\theta} = 1$, so $n\theta$ must be a multiple of $2\pi$. –  Rhys Nov 19 '12 at 19:01
Thank you Rhys. Your words also help! @Rhys –  AnonJ Dec 14 '12 at 16:24

Correction: The $n^{\text{th}}\;$ complex root of unity is represented by $$\text{exp}\left(\frac{2k\pi i}{n}\right), \text{ where}\; n, k\in \mathbb{Z},\;k \geq 0 \text{ and}\; n\geq 1.$$

Recall that the roots of unity are positioned equidistantly along the unit circle. The key is the "$2\pi$" in $2k\pi = k(2\pi)$. It might help to think of the $n^{\text{th}}\;$ roots of unity as forming a regular $n$-gon inscribed in the unit circle (for illustration, see this page).

Typically one considers only the roots $z_i$ (each of which is a solution to $z^n = 1$) for which $$z_i = e^{i\theta}, 0 \leq \theta =\frac{2k\pi}{n} < 2\pi,\; 1\leq i < n;\;k=0, 1, 2, ..., n-1.$$ For our purposes, in terms of finding complex roots of unity, $\;\theta = \theta + 2\pi = \theta + 4 \pi = \cdots$.

For an intuitive overview of complex roots of unity, see YouTube's video.

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k and n needn't be coprime. For instance 4th root of 1 is 1,-1,i,-i. –  Gautam Shenoy Nov 19 '12 at 18:44
Thank you very much. My first journey at math.stackexchange is certainly heartwarming :) –  AnonJ Dec 14 '12 at 15:13
Sure no problem. –  AnonJ Dec 14 '12 at 16:33
Complex style answer, Amy +). I hope to be here as it has been. :-) –  B. S. Sep 15 '13 at 8:22