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Sep 28, 2017 at 21:57 history edited Dierk Bormann CC BY-SA 3.0
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Sep 28, 2017 at 18:30 comment added user90189 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.
Sep 28, 2017 at 15:56 comment added Christian Remling 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).
Sep 28, 2017 at 13:33 history asked Dierk Bormann CC BY-SA 3.0