How to integrate $\int_0^{2\pi} \cos^{10}\theta \mathrm{d}\theta$ using complex analysis. I am asked to evaluate the following integral:
$$\int_0^{2\pi} \cos^{10}\theta \mathrm{d}\theta$$
I am using complex analysis. Setting $z = e^{i\theta}$, I get from Eulers formula:
$$\cos \theta = \frac{1}{2}\left(e^{i\theta} + e^{-i\theta}\right) = \frac{1}{2}\left(z + z^{-1}\right)$$
Now as $\theta$ goes from $0$ to $2\pi$, $z = e^{i\theta}$ goes one time around the unit circle. Therefore the problem is reduced to the following contour integral:
$$\oint_{C} \left(\frac{1}{2}(z + z^{-1})\right)^{10} \frac{dz}{iz}$$ where C is the unit circle.
At this point, I don't know how to move forward. I am pretty sure I am to apply the residue theorem, and the function I am integrating clearly has a pole at $z = 0$. But I don't know how to calculate that residue, since the pole is of the 10th order. Is there another approach I should take, maybe find the Laurent series of the function?
Any help is greatly appreciated!

Hint
Set $$f(z)=\frac{(z+z^{-1})^{10}}{2^{10}iz}=\frac{(z^2+1)^{10}}{2^{10}iz^{11}}.$$
Then, $$f(z)=\frac{1}{iz^{11}}\sum_{k=0}^{10}\binom{10}{k}z^{2k}=...+\frac{1}{i2^{10}z}\binom{10}{5}+...$$
