$\int_{Cr} \frac{100z^{100}+99z^{99}}{z^{100}+z^{99}+1}dz$
I put $f = {100z^{100}+99z^{99}}$ and $g = {z^{100}+z^{99}+1}$
there are 100 roots $z_1, ..., z_{100}$ of $g(z) = 0$
Hence, $\int_{Cr} \frac{f(z)}{g(z)}dz = 2\pi i \sum_{i=1}^{100} \frac{f(z_i)}{g'(z_i)}$
and the answer is $100 \times 2 \pi i = 200\pi i$
Where can I start?
I assume $C_r$ is a sufficiently large circle that encloses all zeros of $f(z)$. To use the residue theorem in this case, you need to show $f(z)$ only has simple zeros, which isn't hard. But $f(z)/g'(z)=z$, so the answer is $2\pi i$ times the sum of all zeros, which is -1 by Viete.