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I need to find the set of complex numbers $z \in \mathbb{C}$ such that $(z-1)^{n}=(z+1)^{n}$

I have found out that $(\frac{z-1}{z+1})^{n}=1$ which reminds me of the roots of unity. But I do not know how to continue from here.

Erik Dz
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    Well, this is true if and only if $\frac{z-1}{z+1}$ is an $n$'th root of unity. So set $\frac{z-1}{z+1}=\zeta_n$, and solve for $z$. – Mastrem Oct 26 '21 at 15:07

2 Answers2

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You already proved that $(z-1)^n=(z+1)^n$ if and only if $\frac{z-1}{z+1}$ is a root of unity.

Now, all you need is to calculate what $z$ must be in order for $\frac{z-1}{z+1}$ to be a root of unity. That shouldn't be hard.

5xum
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The $n$th roots of unity form an equilateral $n$-gon on the unit circle. ($n=7$ is pictured.)

Z1

The linear fractional transformation $z \mapsto \frac{z-1}{z+1}$ maps the imaginary axis onto the unit circle. So the solutions of $(z-1)^{n}=(z+1)^{n}$ are on the imaginary axis (including the point $\infty$).

Z3
(The seventh point is $\infty$.)

GEdgar
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