If $z^3+1=0$ and $\lambda$ is a complex root, prove that $\lambda^2 + 1 = \lambda$.

I have no idea how to start can someone please help me, I will be very grateful.


  • $\begingroup$ Is it $Z^3-1=0$ $\endgroup$ – user171358 Nov 2 '14 at 15:27
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    $\begingroup$ $z^3+1=(z+1)(z^2-z+1)$ $\endgroup$ – Bumblebee Nov 2 '14 at 15:28
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    $\begingroup$ You've asked similar question $\endgroup$ – user171358 Nov 2 '14 at 15:29
  • $\begingroup$ If $n$ is odd, $\forall z\in \mathbb C\left(z^n+1=(z+1)\left(z^{n-1}-z^{n-2}+\ldots +z^2-z+1\right)\right)$ $\endgroup$ – Git Gud Nov 2 '14 at 15:30

$z^3 + 1 = (z+1)(z^2-z+1)$ so the complex root of equation will satisfy the following $z^2-z+1=0$ $\implies z^2+1=z$ which is the required condition


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