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I want to know why if $F(n)$ denote the sum of the primitive $n$th roots of unity in $\mathbb{C}$, and $G(n)$ denote the sum of all complex $n$th roots of unity. Then $G(n)=\sum_{m|n}F(m)$, Please, I need a complete explanation, I understand all the concepts required to the proof but I don't know why I have to take the $m$ such that $m|n$ in the right side of the "$=$"

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Hey, Camilo. You should draw these all down. Circle in the complex plain with the regular $n$-gon starting on $1$. Each one has a rank (e.g. $1$ is a first primitive root, $-1$ is a second, $i$ is a forth..) –  Berci Oct 20 '12 at 1:48
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The $n$th roots of unity is a group $G=<\zeta>$ of order $n$. Every element of $G$, say $\zeta^r$, is a power of $\zeta^{m}$ (where $m=gcd(n,r)$) of order $n/m$, so it is a primitive $n/m$th root of unity

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Every $n$th root of unity is a primitive $m$th root for exactly one $m\mid n$.

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