Root of Unity - Summation

Summation

Let SR(n) be the sum of all the nth roots of unity, primitive or not. Then

\operatorname{SR}(n) = \begin{cases} 1, & n=1\\ 0, & n>1. \end{cases}

For n = 1 there is nothing to prove. For n > 1, it is "intuitively obvious" from the symmetry of the roots in the complex plane. For a rigorous proof, let z be a primitive nth root of unity. Then the set of all roots is given by zk, k = 0, 1, ..., n−1, and their sum is given by the formula for a geometric series:

Let SP(n) be the sum of all the primitive nth roots of unity. Then

where μ(n) is the Mobius function.

In the section Elementary facts, it was shown that if R(n) is the set of all nth roots of unity and P(n) is the set of primitive ones, R(n) is a disjoint union of the P(n):

This implies

Applying the Mobius inversion formula gives

In this formula, if d < n SR(n/d) = 0, and for d = n, SR(n/d) = 1. Therefore, SP(n) = μ(n).

This is the special case cn(1) of Ramanujan's sum cn(s), defined as the sum of the sth powers of the primitive nth roots of unity:

c_n(s)= \sum_{a=1\atop \gcd(a,n)=1}^n e^{2 \pi i \tfrac{a}{n} s} .

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