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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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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? –  copper.hat Aug 20 '12 at 16:35
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Isn't $D$ the unit circle? –  celtschk Aug 20 '12 at 16:56
    
The unit disk in the complex plane is $\{z\in\Bbb C:|z|\le 1\}$. –  Brian M. Scott Aug 20 '12 at 17:04
    
@celtschk yes, it's the unit circle #correction –  pbs Aug 22 '12 at 7:45
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1 Answer

up vote 6 down vote accepted

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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