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Corollary from Cauchy's integral formula
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[QUOTE="STEMucator, post: 4774020, member: 426227"] ##\frac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}] \space (1)## From what I can see, the ##h## in the numerator of this expression will first cancel the ##\frac{1}{h}## term in the integrand. As ##h## gets small, ##(1) → [\dfrac{1}{(ζ−z)(ζ−z)}][\dfrac{n}{(ζ−z)^{n−1}}]##. If you plug in ##A## and ##B## into the formula for ##A^n - B^n##, what is the result if ##h## is small? [B]EDIT:[/B] As another hint, take a look at the first two terms ##A^{n−1} + A^{n−2}B## and the last two terms ##AB^{n−2} + B^{n−1}##. If you add the first two terms when ##h## is small, the result is ##\frac{2}{(ζ-z)^{n-1}}##. If you add the last two terms, what do you get? Notice you're really just adding ##\frac{1}{(ζ-z)^{n-1}}## a total of ##n## times. [/QUOTE]
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Corollary from Cauchy's integral formula
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