Find the cube roots of $-11-2i$. How do I find the roots of   $\sqrt[3]{ - 11 - 2i}$ ?
Tried to use Moivre's theorem, but can not find the solutions by using the polar form:
$z_k=\sqrt{5}[\cos(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})+i.\sin(\frac{\tan^{-1}\frac{2}{11}+2k\pi}{3})]$, for $k=0, 1$ and $2$.
Also resorted to $(a+bi)^3=-11-2i$  and did not have success, because I could not solve 
$a³-3ab² = -11$  and  $3a²b - b³ = -2$.
Solutions at the end of the book are $1+2i$ and $-\frac{1+2\sqrt{3}}{2}+\frac{\sqrt{3}-2}{2}i$ and $-\frac{1-2\sqrt{3}}{2}-\frac{\sqrt{3}-2}{2}i$.

Let's try to find a solution $z=a+bi$ with $a,b \in \mathbb Z$.
We must have $-11=a^3-3ab^2 = a(a^2-3b^2)$ and so $a=\pm 1$ or $\pm 11$.
We must have $-2=3a^2b - b^3 = b(3a^2-b^2)$ and so $b=\pm 1$ or $\pm 2$.
Testing all cases, we see that $a=1$ and $b=2$ works.
Having found one solution $z=1+2i$, the others are $z\omega$ and $z\omega^2$, where $\omega$ is a primitive cubic root of unity:
$$
\omega = \dfrac{-1+\sqrt{3} \, i}{2}
\qquad
\omega^2 = \bar \omega = \dfrac{-1-\sqrt{3} \, i}{2}
$$
