How Square are the Cube Roots of Unity?

Write Unity as $1 = e^{i \cdot 2\pi n}$ where $n$ can be any integer.

$\sqrt[3]{1} = [ e^{i \cdot 2\pi n} ]^{1/3} = e^{i \cdot 2\pi n/3}$

For n=0: $\; \; z = e^{0} = 1$

For n=1: $\; \; z = e^{i \cdot 2\pi /3}$

For n=2: $\; \; z = e^{i \cdot 4\pi /3}$

These are the three unique solutions. All other values of $n$ will give a solution equivalent to one of these three.

Let $z_1 = 1, \; \; \; \; \; \; z_2 = e^{i \cdot 2\pi /3}, \; \; \; \; \; \; z_3 = e^{i \cdot 4\pi /3}$

$z^{2}_2 = [ e^{i \cdot 2\pi /3} ]^{2} = e^{i \cdot 4\pi /3} = z_3$

$z^{2}_3 = [ e^{i \cdot 4\pi /3} ]^{2} = e^{i \cdot 8\pi /3}$ . The phase $8\pi /3 = 2\pi /3 + 2\pi$ which is equivalent to the phase $2\pi /3$ , so

$z^{2}_3 = z_2$

$z^{2}_1 = 1^{2} = 1 = z_1$

So $z^{2}_a = z_b$ when $z_a = 1 = z_b$ , and when $z_a$ is one non-real root and $z_b$ is the other non-real root.

$\large z^{2}_1 = z_1 , \; \; \; \; \; \; z^{2}_2 = z_3 , \; \; \; \; \; \; z^{2}_3 = z_2$