If I wanted to evaluate $\int_{C(0,1)}(z+\frac{1}{z})^{2n}\frac{1}{z}dz$ using the Binomial theorem what would my result be ?
So far I've rearranged the integral until we have $\int\frac{(z^2+1)^{2n}}{z^{2n+1}}dz$
Then by using the binomial theorem on the numerator we obtain
$(1+z^2)^{2n}=\sum_{k=0}^{2n} \binom{2n}{k}(z^2)^k=\binom{2n}{0}+\binom{2n}{1}z^2+...+ \binom{2n}{2n}(z^2)^{2n}$
Then I thought that I should decompose the original integral into a sum of new integrals and evaluate them using Cauchy's integral formula for derivatives, and it seemed like this would yield some fruitful result , but when i applied it I realised that if we take the derivative of $f(z_0)$ 2n times all but the final integral $\binom{2n}{2n}\int \frac{z^{2n+1}}{z^{2n+1}}dz$ will give zero as the result, using this approach.
This last integral then would give $\int dz=\int_{0}^{2\pi} ie^{it}dt$ after parametrisation which also yields zero ?
I don't feel like this is right at all, would anyone have any advice to guide me through where I'm going wrong, your help is much appreciated in advance.
Hint: for $n\in\Bbb Z$, $n\ne -1$ the function $z\longmapsto z^n$ has primitive; for $n = -1$, calculate yourself $$\int_{|z| = 1}\frac1z\,dz$$ with the obvious parametrization $z = e^{i\theta}$ (also works for $n\ne -1$).