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📅 March 1, 2026 👤 scraface

Mar 01, 2026

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z^6 - z^4 + z^2 - 1 = 0
So the roots satisfy either $ z^2 - 1 = 0 $ or $ z^4 + 1 = 0 $.
Roots from $ z^2 - 1 = 0 $:
$ z = \pm 1 $, both real, so imaginary part is 0.
Roots from $ z^4 = -1 $:
Write $ -1 = e^{i\pi} $, so:
z = e^{i(\pi + 2k\pi)/4} = e^{i\pi(2k+1)/4}, \quad k = 0, 1, 2, 3
So the roots are:
$ k=0 $: $ e^{i\pi/4} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} $
$ k=1 $: $ e^{i3\pi/4} = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} $

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