Since any meromorphic function $f:\mathbb C\to\mathbb C$ can be expressed as the quotient of two entire functions, i.e. $f(z) = \frac{n(z)}{d(z)}$ where the zeros of the denominator $d(z)$ are $f$'s singularities, I was wondering how to obtain the latters' Taylor coefficients given the former's Laurent series, i.e.
given the $f_k$ in $f(z) = \sum\limits_{k=-\infty}^\infty f_k z^k$ (the expansion is centered around zero for simplicity), how to obtain the $n_k, d_k$ of $m_k = \sum_{k=0}^\infty m_kz^k, (m\in\{n,d\})$?
Sure enough one can rearrange the fraction to obtain $$\begin{align} d(z)\cdot f(z) &\stackrel!= n(z) \\\Leftrightarrow \sum_{k=-\infty}^\infty \left(\sum_{l=0}^\infty d_l f_{k-l} \right)z^k &\stackrel!= \sum_{k=0}^\infty n_k z^k \\\Rightarrow \sum_{l=0}^\infty d_l f_{k-l} &= \begin{cases} 0 & k<0 \\ n_k & k\ge 0\end{cases} \end{align}$$ i.e. a discrete convolution. But how to solve for the $d_k, n_k$? Are they unique (up to multiplication by a non-vanishing entire function as mrf and Daniel Fischer pointed out)? While the Laurent series does not necessarily converge on all of $\mathbb C$, $n(z), d(z)$ being entire should do so.