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# help

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What is the value of the sum $$\sum_z \frac{1}{{\left|1 - z\right|}^2} \, ,$$ where ranges over all 7 solutions (real and nonreal) of the equation $$z^7 = -1$$?

Mar 25, 2019

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Note: I am really just playing here.  I don't pretend to really know what I am doing.

Mmm

$$z^7=-1$$

One solution for z is  is -1

$$\theta=\pi+\frac{2\pi n}{7}\qquad 0\le n\le6 \qquad n\in Z\\ z=e^{i\theta}=cos\theta +isin\theta\\ z-1=-(1-cos\theta)+isin\theta\\ |z-1|=\sqrt{(1-cos\theta)^2+sin^2\theta}\\ |z-1|=\sqrt{(1+cos^2\theta-2cos\theta)+sin^2\theta}\\ |z-1|=\sqrt{2-2cos\theta}\\ |z-1|=\sqrt2\sqrt{1-cos\theta}\\ \frac{1}{|z-1|^2}=\frac{1}{2(1-cos\theta)}\\$$

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Now

$$Cos\theta = cos(\pi+\frac{2\pi n}{7}\\ Cos\theta = -cos(\frac{2\pi n}{7}\\ so\\ |1-z|^2=2(1+cos(\frac{2\pi n}{7}))\\=2(1+cos(\frac{2\pi *0}{7})),\;\;2(1+cos(\frac{2\pi *1}{7})),\;\; 2(1+cos(\frac{2\pi*2 }{7})),\;\; 2(1+cos(\frac{2\pi *3}{7})),\;\;2(1+cos(\frac{2\pi *4}{7})),\;\;2(1+cos(\frac{2\pi *5}{7})),\;\;2(1+cos(\frac{2\pi *6}{7})),\;\;\\~\\ =2(1+cos(0)),\;\;2(1+cos(\frac{2\pi }{7})),\;\;2(1+cos(\frac{4\pi }{7})),\;\;2(1+cos(\frac{6\pi }{7})),\;\;2(1+cos(\frac{8\pi }{7})),\;\;2(1+cos(\frac{10\pi }{7})),\;\;2(1+cos(\frac{12\pi }{7})),\;\;\\ =4,\;\;2(1+cos(\frac{2\pi }{7})),\;\;2(1+cos(\frac{4\pi }{7})),\;\;2(1+cos(\frac{6\pi }{7})),\;\;2(1+cos(\frac{8\pi }{7})),\;\;2(1+cos(\frac{10\pi }{7})),\;\;2(1+cos(\frac{12\pi }{7})),\;\;\\$$

Put each of these over 1 and add them together and I'd have my answer

The first one is 1/4 so that is right

But I do not think the rest  is right

Mar 25, 2019
edited by Melody  Mar 25, 2019
edited by Melody  Mar 26, 2019
edited by Melody  Mar 26, 2019