I have to find all natural $n$ such that $$I=\int_{|z|=2}\frac{z^n}{z^{10} - 1}dz \neq 0$$
Since all the $10^{\text{th}}$ roots of unity lie in $|z|=2$, by residue theorem we have
$$\int_{|z|=2}\frac{z^n}{z^{10} - 1}dz = 2\pi i \cdot\sum_{k = 0}^{9}\ \text{Res}(f,e^{k\pi i/5})=\frac{2\pi i}{10} \cdot\left(\sum_{k = 0}^{9}z^{n-9}\rvert_{e^{k\pi i/5}}\right) = \begin{cases}0\quad n\not\equiv9\pmod {10} \\ 2 \pi i\quad n\equiv9\pmod {10} \end{cases}$$
Am I correct?
Another way is to use the substitution $w=1/z$. That gives $$I=-\int_{|w|=1/2}\frac{w^{-n-2}}{w^{-10}-1}\,dw =-\int_{|w|=1/2}\frac{w^{8-n}}{1-w^{10}}\,dw.$$ When $n\ge0$ the function $$\frac{w^{8-n}}{1-w^{10}}=\sum_{k=0}^\infty w^{8-n+10k}$$ has a Laurent series about $0$, and the integral is nonzero if its residue (the coefficient of $w^{-1}$) does not vanish. That's the case iff $n$ is a positive integer congruent to $-1$ modulo $10$.