Hello,

I'm trying to find residues for f(z) = $\displaystyle \frac{(z^2 + 1)^n}{z^{n+1}} $ .

A binomial expansion of the numerator gives-

$\displaystyle \frac{1}{z^{n+1}} \sum_{k= 0}^n \binom {n}{k} z^{2k} = \sum_{k= 0}^n \binom {n}{k} z^{2k-n-1}$

And so the coefficient of $\displaystyle \frac{1}{z}$ is obtained when $\displaystyle k = n/2$, Giving a residue: $\displaystyle Res = \binom {n}{n/2}$

My question is: what happens when n is an odd number?