Evaluate $\int_{|z|=3}\frac{dz}{z^3(z^{10}-2)}$ 
$$\int_{|z|=3}\frac{dz}{z^3(z^{10}-2)}$$

There are singularities at $z=0$ and $z^{10}=2\iff z=\sqrt[10]{2}e^{\frac{i\pi k}{5}} \text{  where   } k=0,1,...9$ so we need to find $10$ residues? or have I got it wrong?
 A: Since there are no singularities outside $|z|=2^{1/10}$, the Cauchy Integral Theorem says that
$$
\int_{|z|=3}\frac{\mathrm{d}z}{z^3(z^{10}-2)}
=\int_{|z|=r}\frac{\mathrm{d}z}{z^3(z^{10}-2)}
$$
for all $r\gt2^{1/10}$. As $r\to\infty$, the integral on the right obviously vanishes.
A: This is easy by symmetry, without calculating any residues.
Say $f(z)=\frac1{z^3(z^{10}-2)}$ and let $\omega=e^{2\pi i/10}$. Note that
$$f(\omega z)=\omega^{-3}f(z).$$It follows that the integral is $0$.
Hint regarding the relevant calculations: Say the integral is $I$. Writing slightly informally, $$I=\int_{|z|=3}f(\omega z)\,d(\omega z)
=\int_{|z|=3}\omega^{-3}f(z)\omega\,dz=\omega^{-2}I;$$since $\omega^{-2}\ne1$ this shows $I=0$.
(The "informal" part is the first equality above. To justify it, note that if $t\mapsto\gamma(t)$ is a parametrization of $|z|=3$ then $t\mapsto\omega\gamma(t)$ is also a parametrization.)
A: Hint : Sum of the residues at all singularities including infinity is equal to zero. So it is sufficient to evaluate the singularity of $f$ at $z=\infty$, where $\displaystyle f(z)=\frac{1}{z^3(z^{10}-2)}$ and then apply Cauchy's Residue Theorem.

First line says that, If $f$ has poles at $z=a_i(i=1,2,\cdots ,n)$ then $\displaystyle \sum_{i=1}^n \text{Res}(f;a_i) +\text{Res}(f;\infty)=0$.
A: Actually, that would mean $11$ residues, but since $\operatorname{Res}\left(\frac1{z^3(z^{10}-2)},0\right)=0$…
On the other hand, if $z_k=\sqrt[10]2e^{\frac{k\pi i}5}$ ($k\in\{0,1,\ldots,9\}$), then\begin{align}\operatorname{Res}\left(\frac1{z^3(z^{10}-2)},z_k\right)&=\frac1{{z_k}^310{z_k}^9}\\&=\frac1{10{z_k}^{12}}\\&=\frac{\sqrt[10]2^{-12}}{10}\exp\left(-\frac{12k\pi i}5\right)\\&=\frac{\sqrt[10]2^{-12}}{10}\exp\left(-\frac{12\pi i}5\right)^k.\end{align}So, what you have to sum is a gemetric progression.
