Given that
$$\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}z^{2n+1}$$
verify if there is a multiplicative inverse, that is, a series such that
$$\sin z \sum_{n=0}^{\infty}b_nz^n = 1$$
Is it possible to choose $z_0\in \mathbb{C}$ such that
$$\sin z = \sum a_n(z-z_0)^n$$ and that an inverse exists?
My idea was to compute the series for $\frac{1}{\sin z}$, is there an easier method? Because it says that I need to verifiy if such series exists. Is there a way to find it without even thinking about $\frac{1}{\sin z}$?
You can invert any series that has a positive radius of convergence and non-zero constant term $a_0$.
$$ 1=\sum a_k z^k·\sum b_m·z^m =\sum z^n\sum_{0\le k\le n} a_kb_{n-k} $$ can be recursively solved as $b_0=\frac1{a_0}$ and for $n>0$ $$ b_n=-\frac1{a_0}\sum_{1\le k\le n} a_kb_{n-k} $$
Among other things this means that one can invert the sine series developed around any $z_0$ that is not an integer multiple of $\pi$.