# Math Help - Residue - Solution only for even numbers?

1. ## Residue - Solution only for even numbers?

Hello,
I'm trying to find residues for f(z) = $\frac{(z^2 + 1)^n}{z^{n+1}}$ .
A binomial expansion of the numerator gives-
$\frac{1}{z^{n+1}} \sum_{k= 0}^n \binom {n}{k} z^{2k} = \sum_{k= 0}^n \binom {n}{k} z^{2k-n-1}$
And so the coefficient of $\frac{1}{z}$ is obtained when $k = n/2$, Giving a residue: $Res = \binom {n}{n/2}$
My question is: what happens when n is an odd number?

2. Originally Posted by dudyu
Hello,
I'm trying to find residues for f(z) = $\frac{(z^2 + 1)^2}{z^{n+1}}$ .
A binomial expansion of the numerator gives-
$\frac{1}{z^{n+1}} \sum_{k= 0}^n \binom {n}{k} z^{2k} = \sum_{k= 0}^n \binom {n}{k} z^{2k-n-1}$
And so the coefficient of $\frac{1}{z}$ is obtained when $k = n/2$, Giving a residue: $Res = \binom {n}{n/2}$
My question is: what happens when n is an odd number?
Is there a typo in the question? It looks as though the numerator should be $(z^2 + 1)^n$, not $(z^2 + 1)^2$.

If so, then in the case where n is odd there will be no 1/z term in the expansion, so the residue will be 0.

3. Originally Posted by Opalg
Is there a typo in the question? It looks as though the numerator should be $(z^2 + 1)^n$, not $(z^2 + 1)^2$.

If so, then in the case where n is odd there will be no 1/z term in the expansion, so the residue will be 0.
Yes, there was a typo. thank you!