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Sorry, I know this is a repost, but it won't let me edit.  https://web2.0calc.com/questions/pre-calc-help-please If I say z^n=1, could I do $(\frac{1}{e^{\frac{2\pi i k}{7}}-1})^7=z^7$, then $\frac{1}{(e^{\frac{2\pi i k}{7}}-1)^7}$?  Idk tbh.  Any help is greatly appreciated!! thank you in advanced
 Mar 23, 2023
 #1
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You already posted this here, and it has been answered: https://web2.0calc.com/questions/pre-calc-help-please

 Mar 23, 2023
 #2
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Sorry, I'm not sure if my question is clear then. I already found the solutions: $z= \frac{1}{e^{\frac{2\pi i}{7}}-1}, \frac{1}{e^{\frac{4\pi i}{7}}-1}, \frac{1}{e^{\frac{6\pi i}{7}}-1}, \frac{1}{e^{\frac{8\pi i}{7}}-1}, \frac{1}{e^{\frac{10\pi i}{7}}-1}, \frac{1}{e^{\frac{12\pi i}{7}}-1}$ I am struggling to find how to prove all of those solutions have the same real part and what that real part is. This equation is a 7th root of unity so there has to be some "trick"
 Mar 23, 2023
 #3
avatar+33616 
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Like so:

 Mar 23, 2023
 #4
avatar+132 
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Thank you so much Alan!  I made a slight miscalculation when I tried multiplying by the conjugate.

Saphia1123  Mar 28, 2023

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