Roots of unity and function $\mu$ 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.

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


*

*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 $-1$.

*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?

*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.
