Finding the solution to these equation

Question

Just began a Complex Analysis class and I cannot tell if I'm answering this problem correctly: $z^6 = 1$

What I answered with: $\sqrt[6]{{z}^6}=\sqrt[6]{1}$.

This then implying that: $z = x \cdot 1 + i \cdot 0 \implies x=1$?

Answer 1

$z$ should be powers of the sixth root of unity $\omega_6 = \cos\frac\pi3 + i\sin\frac\pi3 = \frac12 + \frac{\sqrt3}{2} i$. (i.e. $z = \omega_6^k, k = 0,\dots,5$), so that Euler's identity gives $z^6 = \omega_6^{6k} = \cos2\pi k + i\sin2\pi k = 1$.

Answer 2

For a solution not using De Moivre's formula, note that:

$$ z^6-1=0 \;\iff \; (z^3-1)(z^3+1) = 0 \;\iff\; (z-1)(z^2+z+1)(z+1)(z^2-z+1) = 0 $$

Then you get the $2$ real roots $\pm 1$, and $4$ more complex roots from the quadratics $z^2\pm z+1=0\,$.