Looking at a problem of interpolation, I find a Vandermonde type matrix. To be precise I consider the following, let $$A(z)= \sum_{i=1}^p \sum_{j=2}^{n_i+1}\frac{a_{i}^j}{(z-z_i)^j}$$ where the $z_i$ are distinct complex numbers.
I try to find some $a_i^j$, complex numbers, in order to get $$A(c_k)=d_k \hbox{ for } k\in [\vert 1, m\vert]$$ where $m=\sum_i n_i$, $c_k$ are distinct complexe numbers which are distinct from the $z_i$ and $d_k$ are some given complex numbers. Hence it is equivalent to invert the following matrix :
$$ \left(\begin{array}{ccccccc}(c_1-z_1)^{-2} & \dots & (c_1-z_1)^{-n_1-1} & \dots& (c_1-z_p)^{-2} & \dots & (c_1-z_p)^{-n_p-1} \\ \vdots & & & & & & \vdots \\(c_m-z_1)^{-2} & \dots & (c_m-z_1)^{-n_1-1} & \dots & (c_m-z_p)^{-2} & \dots &( c_m-z_p)^{-n_p-1}\end{array}\right) $$
Does anyone have an idea in order to prove that this matix is invertible?