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 $-\frac{1-2\sqrt{3}}{2}-\frac{\sqrt{3}-2}{2}i$.
 A: Here is an elementary way of resolving (without any guesswork) the given system 
$$a^3-3ab^2=-11, b^3-3a^2b=2$$
From the first equation, we obtain
$b^2=\frac{a^3+11}{3a}$.
Hence, the second equation yields
$$b=\frac{2}{b^2-3a^2}=\frac{2}{\frac{a^3+11}{3a}-3a^2}=\frac{6a}{11-8a^3}$$
Plugging this into the first equation, we obtain
$$a^3-3a\frac{36a^2}{(11-8a^3)^2}=-11$$
Setting $a^3=x$ and simplifying, this yields
$$x-\frac{108x}{(11-8x)^2}=-11$$
$$(x+11)(11-8x)^2=108x$$
$$64x^3+528x^2-1923x+1331=0$$
Noticing that $x=1$ is a trivial solution of this one, we can proceed to the following:
$$(x-1)(64x^2+592x-1331)=0$$
So we obtain the solution $a^3=x=1$ which gives $a=1$ and hence $b=2$.
So $z=a+bi=1+2i$ is a solution.
The others can be obtained by computing $\omega \cdot z$ and $\omega^2 \cdot z$ where $\omega$ is the third root of unity.
Remark: These are also the solutions which you obtain when resolving the quadratic in $x$ and then compute the third root of these solutions. But the way using the roots of unity is much nicer..
A: 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:
$$
\omega = \dfrac{-1+\sqrt{3} \, i}{2}
\qquad
\omega^2 = \bar \omega = \dfrac{-1-\sqrt{3} \, i}{2}
$$
A: 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 $1$, $\omega\neq 1$.
