I am attempting to evaluate the integral $$\int_C\left(z+\frac{1}{z}\right)^{2n}\frac{dz}{z}$$ where C is the unit circle centered at the origin.
Using parameterized $z=e^{i\theta}$ and got that $$\int_0^{2\pi}\left(e^{i\theta}+e^{-i\theta}\right)\frac{ie^{i\theta}d\theta}{e^{i\theta}} = 2^{2n}i\int_0^{2\pi}(\cos{\theta})^{2n}d\theta$$ I have been looking at applying Cauchy's Theorem and integral formula, but do not see the connection. Is parameterization the wrong way to approach this?
On
Write the integral as
$$\int_C \frac{(z^2 + 1)^{2n}}{z^{2n+1}}\, dz$$
and apply Cauchy's differentiation formula to $f(z) = (z^2 + 1)^{2n}$. This results in
$$\frac{2\pi i}{(2n)!} \frac{d^{2n}}{dz^{2n}}\bigg|_{z = 0} (z^2 + 1)^{2n}$$
which is $2\pi i$ times the coefficient of $z^n$ in the expansion of $(z^2 + 1)^{2n}$. So it reduces to $2\pi i \binom{2n}{n}$.
Remember, Cauchy's Theorem states that the integral is $i 2 \pi$ times the coefficient of $1/z$ in the integrand. In this case, you simply want the coefficient of $z^0$ from the binomial piece, which by the binomial theorem is $\binom{2 n}{n}$. Thus, the integral is $i 2 \pi \binom{2 n}{n}$.
ADDENDUM
To see where the binomaial coeeficient comes from, recall that the binomial theorem states that
$$\left( z + \frac1{z} \right ) ^{2 n} = \sum_{k=0}^{2 n} \binom{2 n}{k} z^{2 n-k} \frac1{z^k} = \sum_{k=0}^{2 n} \binom{2 n}{k} z^{2 n-2 k} $$
Note that the coefficient of $z^0$, which occurs when $k=n$, is indeed $\binom{2 n}{n}$.