Let $f(z) = \frac{z^5}{1-z^3}$. I want to calculate the following integral: $$ I = \int_C \frac{z^5}{1-z^3}dz, $$ where $C$ is the circle $|z| = 2$.
We have that $$ I = 2\pi i \text{Res}_{z = 1}f(z) =2\pi i \text{Res}_{z=0}\left(\frac{1}{z^2}f(1/z)\right) $$ but I'm struggling with calculating both residues by writing their Laurent series around the desired point. I'm trying to use the geometric series substitution but I'm stuck. Any hints on the substitution?
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I was trying to calculate using residue at infinity, once I did not yet see the formula used by Martin and calculating Laurent series of each singularity is a bit of work. Please correct it if there's some mistake: $$ \frac{1}{z^2}f(1/z) = \frac{1}{z^2}\frac{1/z^5}{1-1/z^3} = \frac{-1}{z^4}\frac{1}{1-z^3} $$ and for $|z^3|<1 \iff |z| <1$ we can write $$ \frac{-1}{z^4}\frac{1}{1-z^3} = \frac{-1}{z^4}\sum_{n=0}^{\infty}(z^3)^n = \sum_{n=0}^{\infty}-z^{3n-4} = -\frac{1}{z^4} -\frac{1}{z}-\cdots $$ so the residue of $\frac{1}{z^2}f(1/z)$ at $z=0$ is $-1$.
As already mentioned in a comment, $$ f(z) = \frac{g(z)}{h(z)} = \frac{z^5}{1-z^3} $$ has poles at all the third roots of unity $z_k = e^{2\pi i k/3 }$, $k = 0, 1, 2$.
The denominator has simple zeros at those points, therefore the residue of $f$ at each $z_k$ is $$ \frac{g(z_k)}{h'(z_k)} = \frac{z_k^5}{-3z_k^2} = -\frac 13 z_k^3 = -\frac 13 \, . $$
Alternatively, $$ \frac{1}{z^2} f(\frac 1z) = \frac{-1}{z^4(1-z^3)} = \frac{-1}{z^4} (1 + z^3 + z^6 + \ldots) = - \frac{1}{z^4} - \frac{1}{z} - z^2 - \ldots $$ for $z \to 0$, so that $$ \operatorname{Res}_{z = 0} \frac{1}{z^2} f(\frac 1z) = -1 \, . $$