# Find solutions for $z^3 + 4\sqrt{3} -4i = 0$ ($n$th roots of complex numbers)

I'm not sure if I have done the last parts of this right.

$z^3=-4\sqrt3+4i$

Let $z^3=w$

So $w=-4\sqrt3+4i = 0$

$|w| = \sqrt{-4\sqrt3^2+4^2}$

$=8$

I have drawn the points $4$ and $-4\sqrt3$ on a complex plane and found out the angle $θ$ (which is from the x axis to the |w| line) to be

= - $\tan^{-1}\frac{4}{-4\sqrt3}$ (the tan is negative because it is in the 2nd quadrant??)

= - $\tan^{-1}\frac{1}{-\sqrt3}$

With the help of the standard triangles I can see this comes to be

= $\fracπ6$

I then have to minus this value from $π$ to find $Arg_z$

$\mathrm{Arg}_z = π-\fracπ6$

= $\frac{5π}6$

Now I put this in exponential form

$w = |w|e^{i(\mathrm{Arg}_z)}$

= $8e^{i(\frac{5π}6)}$

$z^3=w$

$z=w^{1/3}$

= $8^{1/3}e^{i(\frac{5π}6+2kπ)/3}$ when $k = 0, 1, 2$

When $k = 0, z = 8^{1/3}e^{i(\frac{5π}6+2(0)π)/3}$

= $8^{1/3}e^{i(\frac{5π}6)/3}$

= $8^{1/3}e^{i\frac{5π}{18}}$

And the same procedure for when $k = 1, 2$

I haven't done too many of these and was wondering if I am on the right track or not/what changes should I make in my approach?

-
Looks mostly right, except for the very first equality which appears to have an extra $=0$ tacked on the end. I would also introduce the $+2\pi n$ earlier, when I first state the argument of $z$ –  in_wolframAlpha_we_trust Oct 29 '13 at 8:55

One wants to solve $z^3=8v$ where $v=(-\sqrt3+\mathrm i)/2=\exp(\mathrm it)$ with $t=5\pi/6$. Hence the solutions are $z=2\exp(\mathrm it/3+\mathrm i2n\pi/3)$ for some integer $n$, or only for $n$ in $\{0,1,2\}$.
Thus, $z=2\mathrm e^{\mathrm is}$ with $s$ in $\left\{5\pi/18,17\pi/18,29\pi/18\right\}$, or, $z=2(\cos(s)+\mathrm i\sin(s))$.
I'm not sure if I follow the process there, but are you saying your answer for when k(or s in your example) $z=2e^{i\frac{5}{18}}$ and not 8e? –  user88720 Oct 29 '13 at 9:27