Evaluating $\int_{|z|=1}\frac{z^{11}}{12z^{12}-4z^9+2z^6-4z^3+1}dz$ using Rouche's theorem

Question

I am trying to evaluate $$\int_{|z|=1}\frac{z^{11}}{12z^{12}-4z^9+2z^6-4z^3+1}dz$$ Using Rouche's theorem, I know that all zeros of $12z^{12}-4z^9+2z^6-4z^3+1$ are inside $|z|=1$, but computing the residues seems to be challenging. Is there an easy way to proceed that I am not seeing?

Answer 1

$\int_{|z|=1}\frac{z^{11}}{12z^{12}-4z^9+2z^6-4z^3+1}dz=\int_{|z|=R>1}\frac{z^{11}}{12z^{12}-4z^9+2z^6-4z^3+1}dz=\int_0^{2\pi}\frac{i}{12+O(1/R)}d\theta =\frac{\pi}{6}i$ as we let $R \to \infty$

(the first equality crucially uses the fact that the zeroes are inside the unit circle - note also that substitution $z=1/w$ as suggested by @metamorphy works though one has to be careful as it changes the orientation of the unit circle so we get a double minus like in real variables - one from $dz$ and one from getting back to the usual orientation)