Using De Moivre's theorem, the complex roots of $z^3=1$ are:

1. Write $1$ in polar form: $1=\cos 2\pi k+i\sin 2\pi k$ for integer $k$. 2. De Moivre's theorem gives $z_k=\cos\dfrac{2\pi k}{3}+i\sin\dfrac{2\pi k}{3}$. 3. For $k=0$: $z=1$. For $k=1$: $\cos 120^{\circ}+i\sin 120^{\circ}=-\dfrac12+i\dfrac{\sqrt3}{2}$. 4. For $k=2$: $\cos 240^{\circ}+i\sin 240^{\circ}=-\dfrac12-i\dfrac{\sqrt3}{2}$. 5. Solving $z^3=1$ as an ordinary cubic returns only the real root $1$ and misses the other two. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 4: Complex Numbers_