Squaring (around) the circle

Can you find three distinct numbers such that the second is the square of the first, the third is the square of the second, and the first is the square of the third? Sure, probably. Jeez, that was easy ... Assuming you can, what are three such numbers? Ohhhhh, OK, fine, I will actually find them .... you know me so well.

On the face of it, something seems a bit odd here. Let's say the numbers are $a,$ $b$ and $c.$ Then the condition is $b = a^2,$ $c=b^2$ and $a = c^2,$ so combining these conditions we get

$$a = c^2 = b^4 = a^8 \,\, \text{or equivalently} \,\, a^8 - a = a(a^7 -1) = 0.$$ In the real numbers, we would get either $a = 0 (=b=c)$ or $a = 1 (=b=c),$ but since we want unique numbers $a,$ $b,$ $c$ none of these will do.

However, in the complex numbers, squaring a number involves both rotating the plane and squaring the magnitude. If we focus only on unit magnitude (which will remain fixed under squaring), the cyclotomic equation $a^7 - 1 = 0$ has $7$ different solutions

$$a_k = e^{i 2\pi k / 7}, k = 0, 1, \dots, 6.$$ Since $a_0 = 1$ has already been ruled out, we have six possible, distinct solutions $$\begin{align*}a_k &= e^{i 2\pi k / 7} = \cos (2\pi k/7) + i \sin (2\pi k/7),\\ b_k &= e^{i 4\pi k / 7} = \cos (4\pi k/7) + i \sin (4\pi k/7),\\ c_k &= e^{i 8\pi k / 7} = \cos (8\pi k/7) + i \sin (8\pi k/7),\end{align*}$$ for $k = 1, \dots, 6.$