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This question already has an answer here:

Let $z$ be a complex number, and let $n$ be a positive integer such that $z^n = (z + 1)^n = 1$. Prove that $n$ is divisible by 6.


I feel like $|z|=1,$ but I don't know how to prove it. Even if I found that, how would it help? Solutions are greatly appreciated. Thanks in advance!

*I believe there's multiple ways to solve this question. And I would like to solve it using the one without much calculus. I only know up to intermediate algebra.

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marked as duplicate by lab bhattacharjee algebra-precalculus Jan 15 '17 at 3:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @labbhattacharjee wow, I have to wonder how you find duplicates like that sometimes. $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 22:24
  • $\begingroup$ @SimpleArt, I was one of those who posted the answers:) $\endgroup$ – lab bhattacharjee Jan 16 '17 at 2:22
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Since $|z|^n = |z^n| = 1$, $|z| = 1$. Similarly, $|z+1| = 1$. Consider the circles of radius $1$ centred at $0$ and $-1$. Where do they intersect?

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  • $\begingroup$ This looks pretty in my head. :D $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 2:43
  • $\begingroup$ Nice geometric hint! Geometry is my favorite math topic. ;) $\endgroup$ – Regina Dea Jan 15 '17 at 22:55
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Clearly $|z|=1$ since $z^n=1$.

Likewise, $|z+1|=1$.

Combining these, we get $z=-\frac12\pm\frac{\sqrt3}2$, and with some basic inspection, it follows that these are sixth roots of unity where $(z+1)^3\ne1$ and $(z+1)^2\ne1$, so for $z^n=1$ to be true, $n$ must be a multiple of $6$.

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  • $\begingroup$ Please don't copy other people's answers. $\endgroup$ – Tdonut Jan 15 '17 at 2:45
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    $\begingroup$ ^ WTF @Tdonut?, this answer was posted before the other one $\endgroup$ – CIJ Jan 15 '17 at 2:46
  • $\begingroup$ @Tdonut I didn't copy anyone's answer >.> Not to mention, short time spans between similar answers are fine IMO. $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 2:46
  • $\begingroup$ Oh well, I need to sleep, but I think these downvotes are fairly unreasonable. $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 2:49
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Taking norm for both equations lead to: $|z| = |z+1| = 1$. Let $z = x+iy\implies x^2+y^2 = 1 = (x+1)^2+y^2\implies 2x+1 = 0 \implies x = -\dfrac{1}{2}, y = \pm \dfrac{\sqrt{3}}{2}\implies z = e^{i\frac{2\pi}{3}}, z+1 = e^{i\frac{\pi}{3}}\implies \cos(\dfrac{2\pi n}{3})+i\sin(\dfrac{2\pi n}{3}) = 1 = \cos(\dfrac{\pi n}{3})+i\sin(\dfrac{\pi n}{3})\implies \cos(\dfrac{\pi n}{3}) = 0\implies \dfrac{\pi n}{3} = 2\pi k\implies n = 6k\implies 6 \mid n$

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