Calculating a complex integral through residues on the Riemann sphere

Question

I would like to know if the procedure for calculating this integral is correct, or where I make some mistakes. I want to calculate $\int_{\gamma_2} f(z)dz=\int_{\gamma_2} \frac{z^9}{z^{10} -1} dz$, where $\gamma_2$ is the circle with radius 2, considering positive direction. I notice that all poles are within the curve $\gamma_2$. Now, I consider the integral over the Riemann sphere, so I calculate the residue in $\infty$, which is equal to the opposite of the residue in 0 of the function $\frac{f(1/z)}{z^2}$. With a simple calculus, we find that $Res(f, \infty)= -1$. Now, as the sum of all residues over the Riemann sphere equals 0, we can say that the sum of the other residues (the tenth roots of unity) is equal to 1. Then the integral should be equal to $ 2 \pi i$. Is it correct?

Answer 1

I think it is right.
if $\infty$is a isolated singularity, $$\frac{1}{2 \pi i} \int_{\Gamma^{-}} f(z) d z(\Gamma:|z| \rightarrow \rho>r, 0 \leqslant r<\rho<+\infty)$$ $$\operatorname{Res}_{z=\infty} f(z)=-c_{-1}$$ if $f(x)$ have the finite singularity in $\mathrm{C}_{\infty}$,$a_{1}, a_{2}, \cdots \cdot a_{n}, \infty$ $$\sum_{k=1}^{n} \operatorname{Res} f(z)+\operatorname{Res}_{z=\infty} f(z)=0$$ and you can use this to calculate $\operatorname{Res}_{z=\infty} f(z)$ $$\operatorname{Res}_{z=\infty} f(z)=-\operatorname{Res}_{t=0}\left[f\left(\frac{1}{t}\right) \frac{1}{t^{2}}\right]$$ $$\operatorname{res}\left(\frac{z^{9}}{z^{10}+1},\{z, \infty\}\right) = -1$$ https://www.wolframalpha.com/input/?i=Residue%5Bz%5E9%5C%28z%5E10%2B1%29%2C+%7Bz%2C+%5Cinfty%7D%5D