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

February 22, 2026

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

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:

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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