Sorry about the delay. I had a hiccup with the proof.
Anyway, let's calculate
This problem, like just about all problems with integration over complex numbers, is going to depend on just where the zeros of the polynomial in the denominator are. I don't know about you but I would have some slight difficulty in finding the 99 zeros of the denominator. But we may make an appeal to Rouche's theorem. Since we want to integrate over the circle |z| = 1 let's see what we can do on this boundary.
ala Rouche: on (most of) the disk , so all the zeros of lie within the unit circle. (Here's where my glitch occurred: The zeros of the denominator actually lie in a washer on the unit disk. The point is that none of the zeros occur outside the unit disk.)
Now for the fun part. We get the same result integrating over |z| = 1 as we do for any area that encloses |z| = 1 since there are no singularities outside of the unit circle. So let's integrate over the circle |z| = R, where R is very large. In fact let's go for broke and take the limit as R goes to infinity:
Now, the integrand only has singularities inside the unit circle and the boundary (at R) is continuous so we may switch the limit and the integral. I'm also going to let and integrate over
It looks awful, but upon taking the limit we see that only the first term in the denominator survives, so putting the "intermediate step" in for clarity: