1. ## solving this integral

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

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
?

2. didd i solved it correctly
??
did i written the formulas correctly

3. I didn't follow what you did. Why are you messing with $f\left(\frac1z\right)$?

4. residue of infinity

5. Originally Posted by transgalactic
residue of infinity
Why? You just need to find the residue at $0$ and two times the tenth roots of unity.

6. its z in power of 10
thats 10 roots

look at my way is it correct