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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

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Every advice or way to solve will greatly help me!

Idan Daniel
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  • What does $\sqrt[3]{z^3-1}$ mean? – José Carlos Santos Jul 23 '19 at 15:42
  • What does $\square$ mean? – MPW Jul 23 '19 at 15:47
  • the box above the integral doesn't mean anything, I just couldn't delete it where i wrote the formula – Idan Daniel Jul 23 '19 at 15:49
  • Okay thanks, @IdanDaniel – MPW Jul 23 '19 at 15:49
  • @IdanDaniel: I just edit your question for a good look. Feel free to edit again if you wish! – Chinnapparaj R Jul 23 '19 at 15:52
  • The point of José Carlos Santos' remark is that the cube root is not a well-defined function on the complex plane. It is multi-valued, and you need to define which branch you are integrating. – Paul Sinclair Jul 24 '19 at 02:41
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    One other mistake you've made is that there is no such thing as "$\int_{|z|=2} \sqrt[3]{z^3-1}$". What there is, is $$\int_{|z|=2} \sqrt[3]{z^3-1}, dz$$And why this is important is found in your final equation where you completely forgot about that differential and therefore have the wrong integral: $$\int_{|z|=2} z\sqrt[3]{1-\frac{1}{z^3}},dz=\int_{- \pi}^{\pi} 2ie^{i2\theta} \sqrt[3]{1-\frac{1}{8 e^{i3 \theta}}},d\theta$$ – Paul Sinclair Jul 24 '19 at 02:46
  • Paul Sinclair, thanks for the corrections. I edited and wrote what branch we are on. How should I go on from there to solve it? – Idan Daniel Jul 24 '19 at 10:29
  • Unless there is a convention I'm unfamiliar with, the vaue of a function at a single point is not enough to specify a branch. Even if you infer that the cube roots of all positive reals are to be taken as real, you still don't know where the branch cut is made. – Barry Cipra Jul 24 '19 at 12:16

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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.

WimC
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