<s>[INST] 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 $-\frac{1-2\sqrt{3}}{2}-\frac{\sqrt{3}-2}{2}i$.
 [/INST] 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..
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