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I was given a task to solve this equation for $x$:

$$\frac{x-1}{x+1}=z\frac{1+i}{1-i}$$

for a complex number $z$ such that $z^2+z+1=0$. Solving this for $x$ is trivial but simplifying solution using the given condition is what's bothering me. Thanks ;)

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  • $\begingroup$ Notice that this means that $z$ is a cubed root of unity $\endgroup$ – Mathmo123 Jun 22 '14 at 16:14
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You said that solving for $x$ was trivial, so I imagine you were able to get to:

$$x=\frac{1+zi}{1-zi}$$

This is equal to:

$$x=\frac{\left(1+zi\right)^2}{(1-zi)(1+zi)}=\frac{1+2zi-z^2}{1+z^2}$$

Since $1+z+z^2=0$ we can conlude that $1+z^2=-z$ and also that $1=-z-z^2$. We substitute that into the expression to get:

$$x=\frac{(-z-z^2)+2zi-z^2}{-z}=\frac{-z-2z^2+2zi}{-z}=1+2z-2i$$ $$x=1+2z-2i$$

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$$\frac{x-1}{x+1}=z\frac{1+i}{1-i}=z\frac{(1+i)^2}{(1+i)(1-i)}=zi$$

Applying componendo and dividend, $$x=\frac{1+zi}{1-zi}=\frac{(1+zi)^2}{(1-zi)(1+zi)}=\frac{1-z^2+2iz}{-z}=-2i+z-z^2$$ as $z^2+z+1=0\implies z^3=1$

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  • 2
    $\begingroup$ "Solving this for x is trivial but simplifying solution using the given condition is what's bothering me" ($z^2 + z + 1 = 0$) $\endgroup$ – Mathmo123 Jun 22 '14 at 16:19
  • $\begingroup$ @Mathmo123, Is the edited version decent enough? $\endgroup$ – lab bhattacharjee Jun 22 '14 at 16:23
  • $\begingroup$ Works for me :) $\endgroup$ – Mathmo123 Jun 22 '14 at 16:25
  • $\begingroup$ Thanks. I was not familiar with roots of unity until now. $\endgroup$ – Transcendental Jun 22 '14 at 16:31

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