Mobius function and exponential sums

Question

It is easy to show that if $a$ and $n$ are positive integers with $\gcd(a,n)=1$, then $$ \sum_{\substack{z=0 \\ \gcd(z,n)=1}}^{n-1} e^{2\pi i \frac{az}n} = \mu(n). $$ What is the general form of this identity where $\gcd(a,n)=1$ is not assumed?

Answer 1

See Ramanujan sum $g=\gcd(a,n)$ $$\sum_{k=1,\gcd(k,n)=1}^n e^{2i \pi a k/n}=\sum_{k=1,\gcd(k,n)=1}^n e^{2i \pi gk/n}=\frac{\phi(n)}{\phi(n/g)}\sum_{k=1,\gcd(k,n/g)=1}^{n/g} e^{2i \pi k/(n/g)}$$ $$=\frac{\phi(n)}{\phi(n/g)}\sum_{d | \frac{n}g}\mu(d) \sum_{m=1}^{ \frac{n}g/d} e^{2i \pi m/( \frac{n}g/d)}=\frac{\phi(n)}{\phi(n/g)}\sum_{d | \frac{n}g}\mu(d) 1_{\frac{n}g/d=1}=\frac{\phi(n)}{\phi(n/g)}\mu(n/g)$$