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In a question here, the solution given states that $$\zeta=\cos{(\pi/8)}+i\sin{(\pi/8)}$$ is a primitive 8th root of unity. I was under the impression that the primitive roots of unity were given by $$\zeta_n=\cos{(2\pi/n)}+i\sin{(2\pi/n)}$$ which in case wouldn't it be a primitive 16th root of unity? I might have my language messed up here, so I was just looking for clarification.

user26857
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Iceman
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    I agree with you. $\cos(\pi/8)+i\sin(\pi/8)$ is a primitive sixteenth root of unity. – Jack D'Aurizio Jun 12 '15 at 13:50
  • I was lookiing up ways to determine how to find minimal polynomials for sin(k) where k is in degrees, so I came across that answer and the language confused me for a second. I haven't even gone forward to determine whether the question is right or wrong now, but thank you for clarifying. – Iceman Jun 12 '15 at 13:54
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    If $n\gt 2$, there are several primitive roots of unity. A complete list is given by the numbers $\cos(2k\pi/n)+i\sin(2k\pi/n)$, where $k$ ranges over all integers $k$ such that $1\le k\le n-1$ and $k$ and $n$ are relatively prime. The number $\cos(\pi/8))+i\sin(\pi/8)$ is not one of them. – André Nicolas Jun 12 '15 at 14:32
  • That is what I thought, i was under the assumption of the coprimality, but just referenced the first for the problem mentioned. Thanks for the clarification, @AndréNicolas – Iceman Jun 12 '15 at 14:45

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Let me illustrate in case of $4$th roots of unity. Well $x^4=1$ has 4 roots $1, i, -i , -1$ and out of these four $i \; \mbox{and} -i$ have a special property that $i^k \neq 1$ if $k<4$ that means they are strictly fourth roots of unity and no smaller power of these numbers is 1.

user26857
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Miz
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