Solving $z^3 =-1$

Solution 1:

$z=re^{i\theta}$

$z^3 = -1$

$r^3 (e^{3 i \theta})= 1\cdot e^{i\pi}$

Hence, $ r=1$ and $\theta= \frac{i(\pi + 2k\pi) }{3}$ for $k=0, 1,2$

$z= e^{\frac{i\pi}{3}} , e^{i\pi}, e^{\frac{5i\pi}{3}}$

Solution 2:

You can find the complex roots of unity as below. The equation you have solved is not the one that will give you complex roots of unity.

$$z^3=1 \implies z^3-1=0$$ Factor the LHS (using any of a variety of methods): $$(z-1)(z^2+z+1)=0$$ Use the zero-product property: $$z-1=0\implies z=1$$ $$z^2+z+1=0\implies z=\frac{-1\pm i\sqrt{3}}{2}$$ (where we used the quadratic formula above)