# 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.

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It would help if you could state exactly what the question is, rather than some string of symbols. Are you trying to solve for a value $z$ with a certain property? Do you want to express the first expression in terms of $a$ and $b$? What have you tried to do so far? – Lasse Rempe-Gillen Oct 30 '11 at 16:32
I see no equation to solve anywhere... – J. M. Oct 30 '11 at 16:33
Okay, let's be clearer: you want the cube root of $-2+i\sqrt 3$? – J. M. Oct 30 '11 at 17:08
I want to express a and b from upper equality. – Primus Oct 30 '11 at 17:09

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

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

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Thank you for solving this. You are life saver :D – Primus Oct 30 '11 at 17:50
You should accept this as an answer ... – user8250 Oct 30 '11 at 20:41
Maybe replace some cubic roots by sixth roots and check some signs... – Did Nov 4 '11 at 11:46