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Let $n$ be a positive integer, and let $z$ $\in$ $\mathbb{C}$ satisfy the equation $(z-1)^n+(z+1)^n=0.$ The first part of the question asks to show that

$z=\frac{1+w}{1-w}$ for some $w$ $\in$ $\mathbb{C}$ such that $w^n=-1.$

Which is pretty straightforward. The next part asks:

Show that $w\overline{w}=1$

Do I represent $w$ as $a+bi$? Do I put it back into the original equation? There's a similar question that has been asked on the site. But it doesn't address this part of the problem.

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    If $(z-1)^n+(z+1)^n=1$, then $w^n=\left(\frac{z-1}{z+1}\right)^n\ne -1$. Did you mean to write $$(z-1)^n+(z+1)^n=0?$$ – 2017-02-21
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    yes, thank you for pointing that out – 2017-02-21
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    You're welcome. My pleasure. -Mark – 2017-02-21

2 Answers 2

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Note that if $w^n=-1$, then $|w|=1$. Hence, $w=e^{i\phi}$ and $w\bar w=e^{i\phi}e^{-i\phi}=1$. And we are done!

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This actually has nothing to do with your equation. Rather, any number that raised to some real power is $1$ must have modulus $1$.

Observe that $w^{2n}=1$. Writing $w=re^{i\theta}$ reveals that we have $r^{2n}e^{2ni\theta}=1e^{0i}$. Thus $|r|=w\overline{w}=1$