Suppose I have a complex function ($f_{}^{}(z_{}^{})$), analytic in the annular region $\mathcal{C} = \{z : 0\leq r_{}^{}<|z_{}^{}-z_{0}^{}|<R_{}^{} \leq \infty\}$ with the Laurent series expansion around $z_{0}^{}$ given as ,
$$f_{}^{}(z_{}^{})=\sum_{n=-\infty}^{+\infty}a_{n}^{}(z_{}^{}-z_{0}^{})_{}^{n},$$
and also suppose $\frac{1}{f(z)}$ is analytic in the region $\bar{\mathcal{C}}_{}^{} \subseteq \mathcal{C}$ with $z_{0}^{} \in \bar{\mathcal{C}}_{}^{}$ and has the Laurent series expansion around $z_{0}^{}$ given as,
$$\frac{1}{f_{}^{}(z_{}^{})}=\sum_{n=-\infty}^{+\infty}b_{n}^{}(z_{}^{}-z_{0}^{})_{}^{n}.$$
Is there a way to express $b_{n}^{}$'s in terms $a_{n}^{}$'s ?
$\textbf{Note :}$ Please feel free to modify the question by adding more restrictive conditions under which the above question has an answer.
$\textbf{My attempt :}$ Brute force way to multiply above Laurent series expansions of $f(z)$ and $\frac{1}{f(z)}$ (assuming this is meaningful in $\bar{\mathcal{C}}_{}^{}$) and comparing coefficients of $(z-z_{0}^{})_{}^{n}$ to get $\sum_{k=-\infty}^{+\infty}a_{n-k}^{}b_{k}^{} = \delta_{n,0}$ with $n \in \mathbb{Z}$ (how to find solution for this set of simultaneous linear equations? some infinite dimensional version of crammer's method?). (Another approach to solve $\sum_{k=-\infty}^{+\infty}a_{n-k}^{}b_{k}^{} = \delta_{n,0}$ is through Fourier expanding $\{a_{n}^{}\}$ and $\{b_{n}^{}\}$, but this also leads to a version of the above asked problem.)