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.

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    $\begingroup$ This is not what you're asking, but maybe it's worth pointing out that while any rational function with simple poles can be written as in (1) plus a polynomial, this is not true for (2). $\endgroup$ – Christian Remling Sep 28 '17 at 15:56
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    $\begingroup$ Since you already know the terms $b_{ii}$, remove them. You may write $\frac{1}{(z-z_i)(z-z_j)}=\frac{1}{(z_i-z_j)(z-z_i)}+\frac{1}{(z_j-z_i)(z-z_j)}$ and integrate by paths enclosing only two points, so you get $n(n-1)/2$ equations and check the determinant. I hope this helps. $\endgroup$ – user90189 Sep 28 '17 at 18:30

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