I need to find $$\int_{|z|=2} \sqrt[3]{z^3-1}$$ but I'm getting stuck.
What I did: We know that $$\sqrt[3]{z^3-1}=z\sqrt[3]{1-\frac{1}{z^3}}$$ From here, we can write $z=2e^{i \theta}$ and $$\int_{|z|=2} z\sqrt[3]{1-\frac{1}{z^3}}=\int_{- \pi}^{\pi} 2e^{i \theta} \sqrt[3]{1-\frac{1}{8 e^{i3 \theta}}}$$ and I don't really know how to continue. I know that I'm probably not on the right way and will greatly appreciate advice on how to continue and solve integrals such as this one
Every advice or way to solve will greatly help me!
Make the substitution $z\leftarrow z^{-1}$ and rearrange a bit to get it in the form $$\int_{\lvert z \rvert = \tfrac12}-(1-z^3)^{1/3}z^{-3} \mathrm{d}z.$$ Notice that the integrand is meromorphic (with a single pole at $z=0$) on the disc $\lvert z \rvert < \tfrac12$ so you can use Cauchy’s integral formula.