Showing $\mu(n) = \sum_{k} e^{2\pi \frac{k}{n} i}$ using cosine explanation
Let $f(n) = \sum_{(k,n)=1} e^\frac{2\pi ik}{n}$. Show that $f(n) = \sum_{k=1}^{n}\left(\sum_{d\mid (k,n)} \mu(d)e^{\frac{2\pi ik}{n}} \right) $