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I'm not believing the claim (below) that is to be proven true.

Show that if $f$ is entire, not zero on circles of natural radius centred at the origin and $\oint_{|z| = n}\frac{1}{f(z)}\mathrm{d}z \neq\oint_{|z| = n + 1}\frac{1}{f(z)}\mathrm{d}z$, then $f$ is not a polynomial.

I'm thinking that the residue theorem implies that if some roots of $f$ are in the larger circle but not the smaller circle, then the integrals are generally unequal and $f$ could be a polynomial. Where have I misinterpreted the residue theorem?

johnsmith
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  • I agree with you, I do not believe this claim either. – Nathaël Jul 08 '23 at 05:04
  • Take the polynomial $p(z) = z-{3 \over 2}$. Then for $n=1$ the integral is zero, but for $n=2$ the integral is $2 \pi i$. – copper.hat Jul 08 '23 at 05:09
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    Is the inequality supposed to hold for every natural $n$? – Brian Moehring Jul 08 '23 at 05:44
  • @BrianMoehring Yes...that'd be it, right? The degree of a polynomial is finite, so there exists a circle of natural radius that encloses all the roots. Over that circle and the next larger circle, the integrals are equal. Proof by contrapositive would do it, I think. – johnsmith Jul 08 '23 at 06:20
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    Condition missing: exists N such that for all n>N .... – Roland F Jul 08 '23 at 07:41

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