The problem described below appears elementary, but I can't figure out the answer or find it in the literature.  I apologize if I have missed something very basic. 

Let me begin with considering a meromorphic function defined by the following finite sum 
\begin{align*}
f(z)
& = \sum_{i=1}^n \frac{a_i}{z-z_i}
\ , 
\tag*{(1)}
\end{align*}
with $a_i, z_i \in \mathbb C$, all $z_i$ being different.  Then all $a_i$ and $z_i$ are of course uniquely determined by $f$, since $z_i$ are the locations of the poles of $f$, and $a_i$ are their residues.  Once the $z_i$ are known, the $a_i$ can be obtained from the explicit formulas $a_i = \lim_{z \to z_i} (z-z_i) f(z)$ or $a_i = \frac1{2\pi{\rm i}} \oint_{\gamma_i} {\rm d}z\ f(z)$, where $\gamma_i$ is a small enough closed contour around $z_i$.  So far so good.  

Now consider instead a double sum of the form 
\begin{align*}
g(z)
& = \sum_{1\leq i \leq j \leq n} \frac{b_{ij}}{(z-z_i)(z-z_j)}
\ . 
\tag*{(2)}
\end{align*}
Again, the $z_i$ are the locations of the poles of the meromorphic function $g$ and thus uniquely determined by $g$.  

My questions are: 

1. Are also the coefficients $b_{ij}$ still uniquely determined by $g$? 
2. Independently of whether the answer to 1. is yes, can we find an explicit formula for $b_{ij}$ in terms of $g$ and of the pole locations $z_i$, so that (2) becomes valid?  

Note that each $b_{ii}$ is the coefficient of the $(z-z_i)^{-2}$ term in the Laurent expansion of $g$ around $z_i$, and thus determined by $g$.  However, in order to uniquely determine the remaining $n(n-1)/2$ coefficients $b_{ij}$ with $i \neq j$ it does not seem sufficient to focus on the residues of $g$ at $z_i$, since there are only $n$ of the latter.