Prove that: $\oint_c{\dfrac{1}{z^{2n+1}}(z^2+1)^{2n}}=\binom {2n} {n} 2\pi i$

Question

In a mathematical methods problem, where the $c$ is a the unit circle around the origin and in counterclockwise, I need to use a step that I'm not so sure about (Because I don't know how to develop it)

How can I affirm that: $$\oint_c{\dfrac{1}{z^{2n+1}}(z^2+1)^{2n}}=\binom {2n} {n} 2\pi i$$

I know that via Cauchy's integral and the binomial theorem, we get: $$(z^2+1)^{2n}=\sum_{k=0}^{2n}\binom {2n} {n}z^{2k}$$ $$\implies\oint_c{\dfrac{1}{z^{2n+1}}(z^2+1)^{2n}}=\oint_c{\dfrac{1}{z^{2n+1}}(\sum_{k=0}^{2n}\binom {2n} {n}z^{2k})}$$

But I don't know how to apply Cauchy to get a formal justification.

Answer 1

We have from Cauchy's Integral Formula

$$\begin{align} \oint_C\frac{(z^2+1)^{2n}}{z^{2n+1}}dz&=\left.2\pi i \frac{d^{2n}}{dz^{2n}}\left(z^2+1\right)^{2n}\right|_{z=0}\\\\ &=\left.2\pi i \frac{d^{2n}}{dz^{2n}}\sum_{k=0}^{2n}\binom{2n}{k}z^{2k}\right|_{z=0} \tag 1 \end{align}$$

The only non-zero term in $(1)$ comes from the $k=n$ term since for $k<n$, the $2n$'th derivative is zero, while for $k>n$ the $2n$'th derivative is proportional to $z^{2k-2n}\to 0$ as $z\to 0$. Therefore,

$$\bbox[5px,border:2px solid #C0A000]{\oint_C\frac{(z^2+1)^{2n}}{z^{2n+1}}dz=2\pi i\binom{2n}{n}}$$

Answer 2

Employ linearity of the integral (briefly, $\int(f+g)=\int(f)+\int(g)$):

$$\oint_c\frac{1}{z^{2n+1}}\sum_{k=0}^{2n}\binom{2n}{n}z^{2k}\,{\rm d}z=\sum_{k=0}^{2n}\binom{2n}{n}\oint_c z^{2k-2n-1}\,{\rm d}z$$

Do you know how to evaluate $\oint_c z^m\,{\rm d}z$? It's $2\pi i$ if $m=-1$ and $0$ otherwise.