TITLE: The Möbius function is the sum of the primitive $n$th roots of unity.
QUESTION [13 upvotes]: Did you know that the Möbius function $\mu$ is the sum of the primitive nth roots of unity? I want to know about meaning of this. 
This statement is expressed as,
 $$\mu(n) = \sum_{\substack{k=1 \\ (k,n)=1}}^n \exp\left(\frac{2\pi ik}{n}\right).$$

REPLY [5 votes]: A slightly different approach goes like this: introduce
$$\alpha(n) = \sum_{k=1,\; (k,n)=1}^n \exp(2\pi i k/n).$$
Then we have
$$\sum_{k=1}^n \exp(2\pi i k/n) = [[n=1]]
= \sum_{d|n} \sum_{(k,n)=d} \exp(2\pi i k/n)
\\ = \sum_{d|n} \sum_{(k/d,n/d)=1} \exp(2\pi i (k/d)/(n/d))
= \sum_{d|n} \alpha(n/d).$$
Now from the relation
$$\sum_{d|n} \alpha(n/d) = [[n=1]]$$
we may  conclude that $\alpha(n) =  \mu(n)$ by inspection. 
 If this is not sufficient use Mobius inversion to get
$$\alpha(n) = \sum_{d|n} \mu(d) [[n/d=1]]
= \mu(n).$$