Complex Line Integral along the unit circle

Question

Show that: $$\oint_\gamma\frac{1}{z}\left(z+\frac{1}{z}\right)^{2n}\mathrm{d}z=2\pi i\cdot\binom{2n}{n},\quad\text{while > }\gamma=\{z\in\mathbb{C}\,|\,|z|=1\}\,\,(\text{unit circle})$$

So far I get the integrand (using the Binomial Theorem): $$\sum_{k=0}^{2n}\binom{2n}{k}\oint_\gamma\frac{z^{2k}}{z^{2n+1}}\mathrm dz$$ and changed it a little bit: $$\sum_{k=0}^{2n}\binom{2n}{k}\frac{(2n)!}{2\pi i}\oint_\gamma \frac{2\pi i\cdot z^{2k}}{(2n)!z^{2n+1}}$$ by using Cauchy Integral formula we get: $$\sum_{k=0}^{2n}\left(\binom{2n}{k}\cdot\left(\frac{2\pi i}{(2n)!}\cdot\frac{\partial^{(2n)}}{\partial z^{(2n)}}(z^{2k})\right)\right)$$ and that is where I got stuck.

How do I derive $z^{(2k)} 2n$ times (assuming $n$ is bigger than $k$, the answer would be zero and that is incorrect)? or maybe I done it wrong. Please help!

Answer 1

Once we notice that $$\int_\gamma z^m\,dz = \begin{cases} 2\pi i \quad &\text{if } m=-1 \\ 0 \quad &\text{if }m\ne -1\end{cases}$$ the question boils down to finding the coefficient of $z^{-1}$. Since $$z^{-1} (z+z^{-1})^{2n} = z^{-1} \sum_{k=0}^{2n} \binom{2n}{k} z^k (z^{-1})^{2n-k} = \sum_{k=0}^n \binom{2n}{k} z^{2k-2n-1}$$ the relevant term has $k=n$. Hence the coefficient of $z^{-1}$ is $\binom{2n}{n}$.

Answer 2

Here is an approach, parametrizing the contour $\gamma$ as $z=e^{i\theta}$

$$ \oint_\gamma \frac{z^{2k}}{z^{2n+1}}dz= \oint_\gamma {z^{2k-2n-1}}dz=i\int_{0}^{2\pi}e^{2i\theta(k-n)}d\theta = \begin{cases} 2\pi i,\quad k=n \\ 0,\quad k\neq n \end{cases}.$$