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
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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:
- 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:
