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Find the value $$\int_{|z|=2}\frac{1}{z^{100}+1}dz.$$

Im getting struck and my thinking is that by Cauchy theorem it must be zero as $|z| = 2$ is analytic inside ... is my thinking is correct or not pliz verified.

Robert Z
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jasmine
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2 Answers2

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All the function's poles are simple and are the roots of

$$\;z^{100}=-1=e^{\pi i+2k\pi i}=e^{\pi i(1+2k)}\;,\;\;k=0,1,2,...,99$$

Thus, all the residues are within the domain enclosed by the given curve, and thus you can calculate the integral using the residue at infinity theorem (instead of evaluating $\,100\,$ residues...!), so:

$$\frac1{z^2}f\left(\frac1z\right)=\frac1{z^2\left(\frac1{z^{100}}+1\right)}=\frac1{z^2}\left(1-z^{-100}+z^{-200}-z^{-300}+\ldots\right)$$

and we can see the residue is zero, and from here

$$\oint_{|z|=2}\frac{dz}{z^{100}+1}=0$$

DonAntonio
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Hint. No, the integrand is not analytic inside $|z|=2$. Actually there are $100$ (simple) poles in $|z|<2$!! Use the residue theorem: $$\int_{|z|=2}\frac{1}{z^{100}+1}dz=2\pi i\sum_{k=1}^{100}\operatorname{Res}\left(\frac{1}{z^{100}+1},z=w_k\right)$$ where $w_1,w_2,\dots,w_{100}$ are the roots of the equation $z^{100}=-1$. You may also use the residue at infinity's definition.

Robert Z
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  • +1 thanks,,@Robert Z,,,pliz elaborate with explaination, ..i was reading the residue thorem,,,,not getting so much – jasmine Dec 04 '17 at 10:25
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    For the evaluations of the residues see: https://math.stackexchange.com/questions/122786/formula-for-calculating-residue-at-a-simple-pole – Robert Z Dec 04 '17 at 10:32
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    Residue at infinity is much easier, I wonder if it is not the point with the exercise. – A.Γ. Dec 04 '17 at 10:35
  • @ Robert Z ,,,i was trying that i take (z -(-1)^1/100),,here i can not define residue... – jasmine Dec 04 '17 at 10:38
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    Do you know how to evaluate the complex roots of complex number? See for example https://math.stackexchange.com/questions/1870884/find-the-8th-complex-roots-of-81i – Robert Z Dec 04 '17 at 10:42
  • @ Robert Z u just tell me the answer that what is value ? ..after that i will try ,,, – jasmine Dec 04 '17 at 10:44
  • What is the value of what? – Robert Z Dec 04 '17 at 10:46
  • @ Robert Z the value $\int_{|z|=2}\frac{1}{z^{100}+1}dz$? – jasmine Dec 04 '17 at 10:48
  • @ RobertZ im very sorry...ur explaination is not easy to understand as compared to DonAntonio – jasmine Dec 04 '17 at 10:56
  • @lomber lego Do you really understand DonAntonio's answer? BTW I gave you a hint (not a full answer!!) where I also suggested the use of the residue at infinity. – Robert Z Dec 04 '17 at 11:01
  • @RobertZ,,,u have said that |Z|= 2 is not analytics that confused me and other is that i can not able to find residue at infinity from ur given hints,,,,,but anyways im really thankful to u for ur effort and giving me for such valuable time,,,, – jasmine Dec 04 '17 at 11:05
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    Yes, the function $\frac{1}{z^{100}+1}$ is NOT analytic inside $|z|=2$ because it has 100 singularities inside that disc. Moreover I gave you the link https://en.wikipedia.org/wiki/Residue_at_infinity which is exactly the definition used by DonAntonio. – Robert Z Dec 04 '17 at 11:11
  • yes,,,@RobertZ,,,,but after seeing DonAntonio answer i understand ur given answer – jasmine Dec 04 '17 at 11:15
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    This is a mystery to me... – Robert Z Dec 04 '17 at 11:18