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, $$\mu(n) = \sum_{\substack{k=1 \\ (k,n)=1}}^n \exp\left(\frac{2\pi ik}{n}\right).$$
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, $$\mu(n) = \sum_{\substack{k=1 \\ (k,n)=1}}^n \exp\left(\frac{2\pi ik}{n}\right).$$
It is true that the Möbius function $\mu(n)$ is the sum of the primitive $n$th roots of unity.
Perhaps the easiest way to see this is to write $$ \sum_{(k, n) = 1} e^{2\pi i k / n} = \sum_{k = 1}^n \sum_{d \mid (k,n)} \mu(d) e^{2 \pi i k / n} = \sum_{d \mid n} \mu(d) \sum_{\ell = 1}^{n/d} e^{2 \pi i d \ell / n}.$$ We get the first equality by using the property $$ \sum_{d \mid n} \mu(d) = \begin{cases} 1 & \text{ if } n = 1 \\ 0 & \text{ else } \end{cases},$$ and we get the second equality by swapping the orders of summation. In the final expression, the inner summation is a sum of all the $(n/d)$th roots of unity, and thus is zero except in the case when it's trivial, which occurs when $d = n$. So the only surviving term is $\mu(n)$, and we've shown $$ \sum_{(k,n) = 1} e^{2 \pi i k / n} = \mu(n),$$ as you wanted to show. $\diamondsuit$
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
$$\alpha(n) = \sum_{d|n} \mu(d) [[n/d=1]] = \mu(n).$$