$\sum_{n=1}^\infty \frac{\mu (n) z^n}{1-z^n} = z$

Question

Where $\mu$ is the mobius function and $z \in \mathbb{C}$ such that $|z| < 1$. I feel like I have to use the mobius inversion formula for absolutely converging series, setting $F(n) = \frac{z^n}{1-z^n}$ and $f(n) = z$, but I don't get anywhere. The denominator is the real bugger here.

Answer 1

Since $$ \frac{z^n}{1-z^n}=z^n+z^{2n}+z^{3n}+\ldots $$ we have $$ \sum_{n\geq 1}\frac{\mu(n) z^n}{1-z^n}=\sum_{m\geq 1} z^m\sum_{d\mid m}\mu(d) $$ and $\sum_{d\mid m}\mu(d)$ always equals zero unless $m=1$.