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📅 March 13, 2026 👤 scraface

Mar 13, 2026

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So $ z^3 = \omega $ or $ z^3 = \omega^2 $, where $ \omega = e^{2\pi i...
We now solve $ z^3 = \omega $ and $ z^3 = \omega^2 $. Each has three c...
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...
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 /...
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},\...
The imaginary part of $ e^{i\theta} $ is $ \sin \theta $. We compute $...
$ \sin(2\pi/9) = \sin(40^\circ) \approx 0.6428 $

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