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 $-\frac{1-2\sqrt{3}}{2}-\frac{\sqrt{3}-2}{2}i$.

The argument of $-11-2i$ is not $\tan^{-1}\frac{2}{11}$ since the complex vector lies in the third quadrant. So, we have as principal value
$$\arg(-11-2i)=\tan^{-1}\frac{2}{11}-\pi$$
Now, for $k=1$ we have
\begin{align}
z_1&=\sqrt{5}\left[\cos\left(\frac{\tan^{-1}\frac{2}{11}-\pi+2\pi}{3}\right)+i\sin \left(\frac{\tan^{-1}\frac{2}{11}-\pi+2\pi}{3}\right)\right]\\
&=\sqrt{5}\left[\cos\left(\frac{\tan^{-1}\frac{2}{11}+\pi}{3}\right)+i\sin \left(\frac{\tan^{-1}\frac{2}{11}+\pi}{3}\right)\right]\\
\end{align}
Calculator give us
$$z_1=1+2i$$
The another two roots can be found by multiplying $z_1$ by $\omega$ and $\omega^2$, where $\omega$ is a cubic root of $1$, $\omega\neq 1$.
