$\displaystyle \int_{|z|=3}^{nothing}\frac{dz}{z^3(z^{10}-2)}\\$

$\displaystyle f=\frac{1}{z^3(z^{10}-2)}\\$

$\displaystyle f(\frac{1}{z})=\frac{1}{(\frac{1}{z})^3((\frac{1}{ z})^{10}-2)}\frac{z^{13}}{1-2z^{10}}=\\$

$\displaystyle res(f,\infty)= res(\frac{1}{z^2}f(\frac{1}{z}),0)=\frac{1}{z^2}\s um_{n=0}^{\infty}(2z^{10})^n\\$

$\displaystyle res(f,\infty)= res(\frac{1}{z^2}f(\frac{1}{z}),0)=-[res(f,inside|z|=3)+res(f,outside|z|=3)]$

from the sum i get that there is no $\displaystyle z^{-1}$ member in the series

so the coefficient of $\displaystyle z^{-1}$ is zero

so the residiu of infinity is zero

but still all of my singular points are |z|=3

so the integral equals zero

??

did i solved it correctly

did i written every formula regarding the laws of residue correctly here

?