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Something wrong with a proof relating to Laurent coefficients
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[QUOTE="brmath, post: 4533199, member: 486151"] Yes, I kind of assumed f would be analytic; just checking. You are integrating f(z) = 1, so the only Laurent coefficient is ##a_0## = 1, which is as we hope. You may conclude from this that ##| \oint \frac{f(z)}{z-z_0}dz | < M##, but what does that tell you about the Laurent coefficients of f? For example, suppose f(z) = ##z-z_0##, certainly bounded in an annulus around ##z_0##. We have ##a_1## = 1 and all the other coefficients are 0. How would the observation that ##| \oint \frac{f(z)}{z-z_0}dz | < M## tell you that ##a_1## = 1 and all the other coefficients are 0? [/QUOTE]
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Something wrong with a proof relating to Laurent coefficients
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