This question concerns the mapping from the poles of a rational function to the coefficients of its partial fraction decomposition. In general, this mapping is not injective. I want to identify some "minimal" restriction on the domain to make it injective. I conjecture that the restriction described below is sufficient, but I do not see how to approach the proof for higher degrees.

References describing similar work would be appreciated.

We assume that the rational function is the inverse of a polynomial of degree $p\in \mathbb{N}_{\geq 1}$, and we restrict the possible poles as follows.

Let $m\in\mathbb{N}_{\geq 0}$ with $2m\leq p$, and consider sets $\{s_n \}_{n=1}^p$ of the form

$\quad \ s_1=a_1+ib_1$, $ \ \ \ \ \ s_2=a_1 -i b_1$, ...

$s_{2m-1}=a_m + ib_m$, $\ \ s_{2m}=a_m - ib_m$,

$s_{2m+1}=a_{m+1}$, ...

$\quad \ s_{p}= a_{p-m}$

satisfying the following:

  1. $a_n<0$ for every $n$

  2. $a_{n_1}\neq a_{n_2}$ for $n_1\neq n_2$

  3. $b_n>0$ for $1 \leq n \leq m$

Let $\mathcal{S}_{p,m}$ denote the collection of all such sets. For fixed $p$ and $m$, let $\{s_n\}_{n=1}^p \in \mathcal{S}_{p,m}$. Then the conditions above imply that the numbers $s_n^2$ are distinct, so we have the partial fraction decomposition \begin{equation*} \frac{1}{\prod_{n=1}^p (x+s_n^2)} = \sum_{n=1}^p \frac{c_n s_n}{x+s_n^2}. \end{equation*}

If $\{t_n\}_{n=1}^p\in \mathcal{S}_{p,m}$ is distinct from $\{s_n\}_{n=1}^p$ and $\Re(s_n) = \Re(t_n)$ for every $n$, then we can replace $s_n$ by $t_n$ in the sum above to get the rational function \begin{equation*} \frac{Q(x)}{\prod_{n=1}^p (x+t_n^2)} = \sum_{n=1}^p \frac{c_n t_n}{x+t_n^2} . \end{equation*}

Question: Is $Q(x)$ necessarily a non-constant polynomial? For some cases, it is known that the answer is yes, and I would like to prove this in greater generality.

Example: For $p=2$ and $m=1$, we have $s_1=a+ib, s_2 = a-ib$ for some $a<0$ and $b>0$. The partial fraction decomposition is \begin{align*} \frac{1}{ (x+s_1^2)(x+s_2^2)} &= \frac{c_1 s_1}{x+s_1^2} + \frac{c_2 s_2}{x+s_2^2} \end{align*} where \begin{equation*} c_1 = \frac{1}{-4i(a+ib)ab} \quad c_2 = \frac{1}{4i(a-ib)ab}. \end{equation*} If $t_1=a+i\beta$ and $t_2 = a-i\beta$, the polynomial $Q$ is \begin{equation*} c_1 t_1 (x +t_2^2) + c_2 t_2 (x+t_1^2) \end{equation*} and the coefficient of $x$ is \begin{equation*} \frac{1}{4iab} \left( -\frac{a+i\beta}{a+ib} + \frac{a-i\beta}{a-ib} \right). \end{equation*} This term is nonzero for $b\neq\beta$.

Update 1 : We can reformulate the problem as a system of polynomial equations. Suppose \begin{align*} \frac{1}{\prod_{n=1}^p (x+s_n^2)} &= \sum_{n=1}^p \frac{c_n s_n}{x+s_n^2}\\ \frac{1}{\prod_{n=1}^p (x+t_n^2)} &= \sum_{n=1}^p \frac{d_n t_n}{x+t_n^2}\\ \end{align*} for distinct sets $\{s_n\}_{n=1}^p,\{t_n\}_{n=1}^p\in \mathcal{S}_{p,m}$ with $\Re(s_n) = \Re(t_n)$ for every $n$.

Question: Does there exist $\sigma \in \mathbb{R}\backslash\{0\}$ such that $c_n^{-1} = \sigma d_n^{-1}$ for every $n$.

Using Cauchy's integral formula, the coefficients are \begin{equation*} c_ns_n = \frac{1}{\prod_{k\neq n} (-s_n^2 +s_k^2) }, \end{equation*} so \begin{align*} c_n^{-1} &= s_n\prod_{k\neq n} (-s_n^2 +s_k^2) \\ d_n^{-1} &= t_n\prod_{k\neq n} (-t_n^2 +t_k^2) \\ \end{align*}


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