Complex integral question

[nhrock3]

$$\int_{|z|=3}^{nothing}\frac{dz}{z^3(z^{10}-2)}\\$$
$$f=\frac{1}{z^3(z^{10}-2)}\\$$
$$f(\frac{1}{z})=\frac{1}{(\frac{1}{z})^3((\frac{1}{z})^{10}-2)}\frac{z^{13}}{1-2z^{10}}=\\$$
$$res(f,\infty)= res(\frac{1}{z^2}f(\frac{1}{z}),0)=\frac{1}{z^2}\sum_{n=0}^{\infty}(2z^{10})^n\\$$
$$res(f,\infty)= res(\frac{1}{z^2}f(\frac{1}{z}),0)=-[res(f,inside|z|=3)+res(f,outside|z|=3)]$$

from the sum i get that there is no $$z^{-1}$$ member in the series
so the coefficient of $$z^{-1}$$ 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
?

 

[Nino MMP]

Yes, you solved it correctly. The formula for the residue of a function is correct and the result is that the integral is equal to zero since the residue at infinity is zero.