complex numbers - How to solve $2 - sqrt{3}i + z^{3}=0$?

I have tried to solve this "problem" so that i have $z$ switched with $a+bi$. Then after some hours of solving this riddle i have gain an enormous numbers of lines but with no solution.

$2 - \sqrt{3}i + z^{3}=0$

$z=a+bi$

I want to know how much is a and b.

Answer

$z^3=-2+\sqrt{3}i$

Let's express $-2+\sqrt{3}i$ in form of $r(\cos\theta+i\sin\theta)$

$r=\sqrt{4+3}=\sqrt{7}$

$\theta=\arctan(\frac{\sqrt{3}}{2})+\pi$

so we may write that:

$z^3=\sqrt{7}(\cos\theta+i\sin\theta)$

Now if you apply following formula you can calculate $a$ and $b$:

$z=\sqrt [3] {7}(\cos(\frac{\theta+2k\pi}{3})+i\sin(\frac{\theta+2k\pi}{3}))$ ,where $k \in \mathbb{Z}$

so: $a=\sqrt [3] {7}\cos(\frac{\theta+2k\pi}{3})$ and $b=\sqrt [3] {7}\sin(\frac{\theta+2k\pi}{3})$