I am reading about Moebius inversions, and I am given the following claim:
$\sum_{d=1}^{\infty} f(z^d) = g(z)$ with $g(z) = O(z)$ at $z=0$, implies that $f(z) = \sum_{d=1}^{\infty} \mu(d)g(z^d)$ by directly applying the Moebius inversion to the relationship on the coefficients.
The relationship I see on the coefficients is $g_n = \sum_{d|n} f_d$, but if I apply the Moebius inversion relation, I would get that $f_n = \sum_{d|n} g_d\mu(n/d)$
I don't see how to get the result from here.
Write your coefficient relation
$$f_n = \sum_{d\mid n} \mu(d) g_{n/d}.$$
Then multiply with $z^n$:
$$f_n z^n = \sum_{d\mid n} \mu(d) g_{n/d}(z^d)^{n/d}$$
and sum:
$$f(z) = \sum_{n=1}^\infty f_n z^n = \sum_{n=1}^\infty \sum_{d\mid n} \mu(d) g_{n/d}(z^d)^{n/d} = \sum_{d=1}^\infty \mu(d)\sum_{d\mid n} g_{n/d}(z^d)^{n/d} = \sum_{d=1}^\infty \mu(d)g(z^d).$$