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I'm having difficulty understanding how to prove that the primitive roots of unity are in fact dense on the unit circle. I have the following so far:

The unit circle can be written $D=\{x\in\mathbb{C}:|x|=1\}$. The set of primitive $m$-th roots of unity is $A_m=\{\zeta_k:\zeta_k^m=1,\zeta_k\text{ is primitive}\}$.

Hence, the set of all primitive roots $A$ is given by the union of $A_m$ over $m=1,2,3,\ldots$. But I can't seem to get started on how to prove that $A$ is dense in $D$.

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    $\begingroup$ Possibly you could show that $\{\frac{m}{n} | \gcd(m,n) = 1\}$ is dense in $\mathbb{R}$, and use continuity of $x \mapsto e^{2 \pi i x}$ to finish? $\endgroup$ – copper.hat Aug 20 '12 at 16:35
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    $\begingroup$ Isn't $D$ the unit circle? $\endgroup$ – celtschk Aug 20 '12 at 16:56
  • $\begingroup$ The unit disk in the complex plane is $\{z\in\Bbb C:|z|\le 1\}$. $\endgroup$ – Brian M. Scott Aug 20 '12 at 17:04
  • $\begingroup$ @celtschk yes, it's the unit circle #correction $\endgroup$ – Pixel Aug 22 '12 at 7:45
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To show the primitive roots of unity are dense in the unit circle, we must show that if we pick any point on the unit circle and any $\epsilon>0,$ there is some primitive root of unity with distance less than $\epsilon$ from the chosen point.

The easy way is to specialize $m$ to be prime - then every root of unity other than $1$ is primitive. Since the roots will be distributed evenly along the circle, if we pick a large enough prime we can make the distance between any two adjacent primitive roots less than $\epsilon.$ Then certainly if we pick any point on the unit circle, it will be within $\epsilon$ from a root.

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