Exercise 4.3.28* (The sum of the primitive $n$-th roots of unity is equal to $\mu(n)$)

Proof. By Problem 25, the primitive $n$-th roots $\zeta$ of unity are $\zeta =e^{2\mathit{\pi ia}∕n}$, where $0\le a<n$ and $a\wedge n=1$, and by Problem 26,

If we expand ${\Phi }_n(x)$, then, writing $\delta =\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_n(x))=\varphi (n)$,

where the sum of the roots of ${\Phi }_n(x)$ is given by

It remains to compute the coefficient ${\sigma }_1$ of $x^{\delta -1}$ in ${\Phi }_n(x)$, using (1).

To find an asymptotic expansion of ${\Phi }_n(z)$ in a neighborhood of $\infty$, we compute the Taylor serie (limited development) of $z^{\delta }{\Phi }_n(1∕z)$ of order $1$ in a neighborhood of $0$.

First

thus

Next, by (1),

We know that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}\mathit{d\mu }(n∕d)=\varphi (n)=\delta$ (by the Möbius inversion formula applied to $n={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}\varphi (d)$, or the comparison of the degrees in (1)). Therefore

If $d>1$ then $1-z^d=1+o(z)$, and if $d=1$, $1-z^d=1-z=1-z+o(z)$. Therefore

Hence

The comparison of (3) and (4) gives by the unicity of the coefficients in an expansion,

Therefore (2) becomes

If $n$ is a positive integer, the sum of the primitive $n$-th roots of unity is equal to $\mu (n)$. □