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📅 May 4, 2026 👤 scraface

May 04, 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:

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