<s>[INST] Complex Numbers - Converting to Polar form I understand the basics of converting to polar form but I have just come across a question that I haven't seen before.
Usually the complex number is expressed as $z=a+bi$, but this time the complex number I was given is $z^3=-4+4{\sqrt3}i$. 
Do I need to somehow remove the 3 power? 
Do I just simply use my usually formulas and ignore the power? 
Thanks.
EDIT: I thought I should add additional information. I need to convert to polar form, as I will then use Moivre's rule to calculate the roots of the complex number. 
I found the polar form for the RHS, $z=8(\cos2.09+i\sin1.05)$. Do I need to cube root this to find the polar form of the original complex number?
Moivre expression is in the picture attached Moivre Expression
where $n=3$ and $k=0,1,2$ I am to find the roots $z0, z1, z2$
 [/INST] Now $z^3=8(\cos \frac{2\pi}{3}+i \sin \frac{2 \pi}{3})$: using DeMoivre's formula we get 
$$
z_0=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 0 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 0 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{2\pi}{9}+i \sin \frac{2 \pi}{9}\right)\\
z_1=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 1 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 1 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{8\pi}{9}+i \sin \frac{8 \pi}{9}\right)\\
z_2=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 2 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 2 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{14\pi}{9}+i \sin \frac{14 \pi}{9}\right)
$$
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