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I need to prove that for each positive integer $n$ the sum of the primitive $n$th roots of unity in $\mathbb{C}$ is $\mu(n)$, where $\mu$ is the Möbius function.

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So what's the question? – Patrick Da Silva Oct 20 '12 at 0:02
Yes...or not: what's your question? ¿Cuál es tu pregunta? – DonAntonio Oct 20 '12 at 0:03
I need one proof to the given proposition – Elmo goya Oct 20 '12 at 0:11
Well explained... not wikipedia... please!!! – Elmo goya Oct 20 '12 at 0:33
You can try to prove it for primes, then $p^n$, then use multiplicativity. – Berci Oct 20 '12 at 0:36
up vote 5 down vote accepted

Do you know $$\sum_{d\mid m}\mu(d)=1{\rm\ if\ }m=1,\,\,=0{\rm\ else}$$ The sum of the primitive $n$th roots of unity is $$\sum_{\gcd(k,n)=1}e^{2\pi ik/n}=\sum_1^n\sum_{d\mid\gcd(k,n)}\mu(d)e^{2\pi ik/n}=\sum_{d\mid n}\mu(d)\sum_0^{(n/d)-1}e^{2\pi idk/n}$$ The inner sum os the sum of all the $m$th roots of unity where $m=n/d$, so it's zero except for $d=n$ when it's $1$. So, the original sum evaluates to $\mu(n)$.

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Let $\theta$ denote the first $n$th primitive root: $\theta:=e^{2\pi i/n}$.

  1. If $n=p$ is prime, $\mu(p)=-1$ and each $0\ne a<p$ is relatively prime to $p$, so this $\theta^{a}$ is primitive $p$th root. The sum of all $n$th roots is always $0$ (because if we multiply it by $\theta$, it doesn't change). So we miss only the $\theta^0=1$, hence the sum is $-1$.
  2. If $n=p^k$ ($k\ge 2$), then $\mu(n)=0$ and exactly the $p\cdot a$ elements have common divisor with $n$, so $$\sum_{\theta^u\text{ prim.root}}\theta^u=\sum_{u\ne a\cdot p}\theta^u = \sum_{u=0}^{n-1}\theta^u-\sum_{v=0}^{\frac np-1} \theta^{pv} $$ Can you continue?
  3. You also need to show that both functions in question are multiplicative, i.e., whenever $\gcd(a,b)=1$, we have $$\mu(ab)=\mu(a)\cdot\mu(b) $$ and same for the other function.

From these the proposition follows.

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