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Stop Waiting! Find Moneypass ATMs Near Me Today! - SUpost

📅 July 6, 2026 👤 scraface

Jul 06, 2026

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📌 So \( z^3 = \omega \) or \( z^3 = \omega^2 \), where \( \omega = e^{2\pi i/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \), and \( \omega^2 = e^{-2\pi i/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \), both primitive cube roots of unity.

📌 We now solve \( z^3 = \omega \) and \( z^3 = \omega^2 \), each giving 3 roots.

📌 Let’s find the roots of \( z^3 = \omega \). Write \( \omega = e^{2\pi i/3} \), so:

📌 z = \omega^{1/3} \cdot e^{2\pi i k / 3} = e^{2\pi i/9} \cdot e^{2\pi i k / 3} = e^{2\pi i (1/9 + k/3)} = e^{2\pi i (1/9 + 3k/9)} = e^{2\pi i (3k+1)/9}, \quad k = 0,1,2

📌 Similarly, for \( z^3 = \omega^2 = e^{-2\pi i/3} \):

📌 z = e^{-2\pi i/9} \cdot e^{2\pi i k / 3} = e^{-2\pi i (1 - 3k)/9} = e^{2\pi i (2/9 + k/3)} = e^{2\pi i (2 + 3k)/9}, \quad k = 0,1,2

📌 So all roots have arguments:

📌 From first set: \( \theta = \frac{2\pi(3k+1)}{9} \), \( k=0,1,2 \) → \( \frac{2\pi}{9}, \frac{8\pi}{9}, \frac{14\pi}{9} \)

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