Complex Numbers - Converting to Polar form

Question

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 $z0, z1, z2$

Answer 1

Now $z^3=8(\cos \frac{2\pi}{3}+i \sin \frac{2 \pi}{3})$: using DeMoivre's formula we get $$ z_0=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 0 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 0 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{2\pi}{9}+i \sin \frac{2 \pi}{9}\right)\\ z_1=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 1 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 1 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{8\pi}{9}+i \sin \frac{8 \pi}{9}\right)\\ z_2=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 2 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 2 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{14\pi}{9}+i \sin \frac{14 \pi}{9}\right) $$

Answer 2

You need to write $z^3=r(\cos\theta+i\sin\theta)$. To find $r$, you take the absolute value, and you found $r=8$. Then $$z^3=-4+4\sqrt 3i=8(\cos\theta+i\sin\theta)$$ This means $$\cos\theta=-\frac 12\\\sin\theta=\frac{\sqrt 3}2$$ From here you can find $\theta=\frac 23\pi$. Then, applying Moivre, you get $$z=\sqrt[3] 8\left(\cos\left(\frac{2\pi}9+\frac{2k\pi}3\right)+i\sin\left(\frac{2\pi}9+\frac{2k\pi}3\right)\right)$$ Here $k=0,1,2$, and you can also use $\sqrt[3] 8=2$.

Answer 3

The modulus of $z^3$ is $\sqrt{16(1+3)}=8$, so $$z^3=8\Bigl(-\frac12+\frac{\sqrt 3}2i\Bigr)=8\mathrm e^{\tfrac{2i\pi} 3}$$ Therefore, if $\;z=r\mathrm e^{i\theta}$, we have to solve \begin{cases} r^3=8\enspace (r\in\mathbf R^+)\iff r=2,\\[1ex] 3\theta\equiv \frac{2\pi}3\mod 2\pi\iff\theta\equiv \frac {2\pi}9\mod\frac{2\pi}3. \end{cases} This results in $3$ solutions since the complex exponential has period $2i\pi$, $$z_1=2\mathrm e^{\tfrac{2i\pi}9}, \quad z_2=2\mathrm e^{\tfrac{8i\pi}9}, \quad z_3=2\mathrm e^{\tfrac{14i\pi}9}\;(=2\mathrm e^{-\tfrac{4i\pi}9}).$$