To find the cube roots of
[tex]z=4\sqrt[]{3}+4i[/tex]
First we're going to write this number in polar form:
[tex]z=4(\sqrt[]{3}+i)=8(\frac{\sqrt[]{3}}{2}+\frac{1}{2}i)[/tex]
A number in polar form is written in the following way:
[tex]z=r(\cos \theta+i\sin \theta)[/tex]
By comparing this form with our number, we get the 'r' and the angle from our number:
[tex]\begin{cases}r=8 \\ \theta=\frac{\pi}{6}\end{cases}[/tex]
From this, we get:
[tex](4\sqrt[]{3}+4i)^{\frac{1}{3}}=2(\cos \frac{\pi}{6}+i\sin \frac{\pi}{6})^{\frac{1}{3}}[/tex]
Those 3 roots are:
[tex]\begin{gathered} r_1=2(\cos \frac{\pi}{18}+i\sin \frac{\pi}{18})^{} \\ r_2=2(\cos (\frac{\frac{\pi}{6}+2\pi}{3})+i\sin (\frac{\frac{\pi}{6}+2\pi}{3}))^{} \\ r_3=2(\cos (\frac{\frac{\pi}{6}+4\pi}{3})+i\sin (\frac{\frac{\pi}{6}+4\pi}{3}))^{} \end{gathered}[/tex]
Rewritting those roots:
[tex]\begin{gathered} r_1=2(\cos \frac{\pi}{18}+i\sin \frac{\pi}{18}) \\ r_2=2(\cos (\frac{13\pi}{18})+i\sin (\frac{13\pi}{18}))^{} \\ r_3=2(\cos (\frac{25\pi}{18})+i\sin (\frac{25\pi}{18}))^{} \end{gathered}[/tex]
