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We know that in the set of nth roots of unity there are exactly $n$ distinct complex numbers: $C_n = \{z \in \mathbb{C} : z^{n} = 1 \} = \{e^{i2k\pi/n}, \; k = 0, 1, ...,n-1 \} $ and they lie on the unit circle. Now collect all roots of unity for every $n \in \mathbb{N}$ and consider the set $C = \{C_n, \; n \in \mathbb{N}\}$.

Is the set $C$ the unit circle, that is $\{C_n, \; n \in \mathbb{N}\} = \{z \in \mathbb{C} : |z| = 1 \}$ ?

Obviously $\{C_n, \; n \in \mathbb{N}\} \subseteq \{z \in \mathbb{C} : |z| = 1 \}$, but I didn't manage to prove the other inclusion (if it is true).

Massimo
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  • No. http://math.stackexchange.com/questions/184666/help-proving-the-primitive-roots-of-unity-are-dense-in-the-unit-circle – Nosrati Mar 21 '17 at 13:19
  • Well, no: of course not. There are many element in $;S^1=$ the unit circle, that are not roots of unity, for example $;e^{i/\pi^2};$ ... – DonAntonio Mar 21 '17 at 13:20

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