Using function,

$f(z)=-\left(\frac{z-1}{z+1}\right)^2\frac{2n/z}{z^{2n}-1}$

should give residues as follows:

$ \begin{align} \ &\left(\operatorname*{Res}_{z=0}(f(z))+\operatorname*{Res}_{z=-1}(f(z))\right) = &\left(2n-\frac23(2n^2+1)\right) \end{align} $

However when I use mathematica residue function:

Residue[-((z - 1)/(z + 1))^2 (((2 n)/z)/(-1 + z^(2 n))), {z, -1}]
Residue[-((z - 1)/(z + 1))^2 (((2 n)/z)/(-1 + z^(2 n))), {z,  0}]

I do not get the same result.

What do I have to do to get mathematica to work out these residues?

Using the function,

$f(z)=-\left(\frac{z-1}{z+1}\right)^2\frac{2n/z}{z^{2n}-1}$,

should give residues as follows:

$ \begin{align} \ &\left(\operatorname*{Res}_{z=0}(f(z))+\operatorname*{Res}_{z=-1}(f(z))\right) = &\left(2n-\frac23(2n^2+1)\right) \end{align} $

However when I use the Mathematica Residue[] function:

Residue[-((z - 1)/(z + 1))^2 (((2 n)/z)/(-1 + z^(2 n))), {z, -1}]
Residue[-((z - 1)/(z + 1))^2 (((2 n)/z)/(-1 + z^(2 n))), {z,  0}]

I do not get the same result.

What do I have to do to get Mathematica to work out these residues?