Who closed this question? the similar one is actually different, if you see the answer.
The mobius function $ \mu (n)$, is a function with property:
$$\mu(1) = 1$$
$$\mu(n) = 0$$ if $n$ is of the form $p^{2} \times (...)$
$$\mu(n) = (-1)^{j}$$ if $n$ is exactly product of $j$ distinct primes
For example: $\mu(16) = 0$, $\mu(15)=1$, $\mu(14)=1$, $\mu(13)=-1$.
My question is, how to show that it can be written using the complex form:
$$ \mu(n) = \sum_{k} e^{2\pi \frac{k}{n} i}$$, for $1 \le k \le n$ such that $gcd(k,n)=1$
Attempt:
For the complex part, it will be $\sin \left(2 \pi \frac{k}{n} \right)$, and this will be zero because if $k$ is such that $gcd(k,n)=1$ then we must also have $gcd(n-k,n)=1$, and both $\sin(2 \pi k/n)$ and $\sin((n-k)2\pi/n)$ will cancel each other.
But how to see the relation of the cosine?
