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$\int z^2 \log [(z+1)/(z-1)] dz$ taken around circle $|z|=2$

I am taking residues at $\pm 1$.

This gives me 0 as the value of integral. Is this correct.

How do I modify the integral to get value over half the circle?

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    $\begingroup$ The residue at $z=\pm 1$ does not make sense. You should take the residue at infinity. $\endgroup$ – Kemono Chen Dec 26 '18 at 5:46
  • $\begingroup$ I need to take residues at poles right and those poles which lie inside the |z|=2 $\endgroup$ – Sonal_sqrt Dec 26 '18 at 5:47
  • $\begingroup$ Yes. $\pm1$ are not poles, they are branch points. $\endgroup$ – Kemono Chen Dec 26 '18 at 5:47
  • $\begingroup$ But Cauchy residue theorem requires that poles be inside |z|=2, which as I see it there are no such poles. Then we can't apply residue theorem can we? $\endgroup$ – Sonal_sqrt Dec 26 '18 at 5:58
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$$\int_C f(z)dz=-2\pi i \operatorname{Res}_{z=\infty}f(z)$$ But we have $$f(z)=2z+\frac23z^{-1}+o(z^{-1})$$ as $z\to\infty$, therefore the integral equals $\frac43 \pi i$. $f$ is not meromorphic in $|z|<2$ , so we can not apply residue theorem inside the circle, but we can apply it outside the circle.

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  • $\begingroup$ But Cauchy residue theorem requires that poles be inside |z|=2, which as I see it there are no such poles. Then we can't apply residue theorem can we? $\endgroup$ – Sonal_sqrt Dec 26 '18 at 6:04
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    $\begingroup$ @Piyush Cauchy residue theorem requires meromorphicity. The function is not even meromorphic in that circle. $\endgroup$ – Kemono Chen Dec 26 '18 at 6:17

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