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Home » Discover the Scandal Behind Vilgefortz: Hollywood Would Never Show This! - Midis

Discover the Scandal Behind Vilgefortz: Hollywood Would Never Show This! - Midis

📅 March 12, 2026 👤 scraface

Mar 12, 2026

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📌 So $ z^3 = \omega $ or $ z^3 = \omega^2 $, where $ \omega = e^{2\pi i / 3} = \frac{-1 + i\sqrt{3}}{2} $, a primitive cube root of unity. The other root is $ \omega^2 = e^{-2\pi i / 3} $.

📌 We now solve $ z^3 = \omega $ and $ z^3 = \omega^2 $. Each has three cube roots.

📌 First, write $ \omega = e^{2\pi i / 3} $, so the cube roots are:

📌 z = \omega^{1/3} = e^{2\pi i / 9},\ e^{2\pi i (1+3)/9} = e^{8\pi i / 9},\ e^{2\pi i (1+6)/9} = e^{14\pi i / 9} \quad \text{(mod } 2\pi).

📌 Similarly, $ z^3 = \omega^2 = e^{-2\pi i / 3} $, so roots:

📌 z = e^{-2\pi i / 9},\ e^{-8\pi i / 9},\ e^{-14\pi i / 9} = e^{4\pi i / 9} \quad \text{(since } -14\pi/9 \equiv 4\pi/9 \mod 2\pi).

📌 So the six roots are:

📌 e^{2\pi i / 9},\ e^{8\pi i / 9},\ e^{14\pi i / 9},\ e^{-2\pi i / 9},\ e^{-8\pi i / 9},\ e^{4\pi i / 9}.

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