Given the complex number : -64 i
To find the cube roots, we will use the formula:
[tex]\sqrt[3]{r}(\cos \frac{\theta+2\pi k}{3}+i\sin \frac{\theta+2\pi k}{3})[/tex]
k = 0 , 1 , 2
So,
[tex]r=-64i=64\angle270[/tex]
So,
[tex]\sqrt[3]{64}=4[/tex]
and the angles will be :
[tex]\begin{gathered} \frac{\theta+2\pi k}{3}=\frac{\theta+360k}{3} \\ k=0,\frac{\theta+360\cdot0}{3}=\frac{270}{3}=90 \\ \\ k=1,\frac{\theta+360\cdot1}{3}=\frac{270+360}{3}=210 \\ \\ k=2,\frac{\theta+360\cdot2}{3}=\frac{270+720}{3}=330 \end{gathered}[/tex]
so, the cube roots of the -64i are:
[tex]\begin{gathered} 4(\cos 90+i\sin 90) \\ 4(\cos 210+i\sin 210) \\ 4(\cos 330+i\sin 330) \end{gathered}[/tex]
So, the answer is option C
