<s>[INST] How to find $(-64\mathrm{i}) ^{1/3}$? How to find $$(-64\mathrm{i})^{\frac{1}{3}}$$
This is a complex variables question.
I need help by show step by step.
Thanks a lot.
 [/INST] If you transform $-64i$ to polar form, you get $r=\sqrt{0+(-64)^2}=64$ and $\theta=-\pi/2$.
Then you have $$(-64i)^{1/3} = r^{1/3}\cdot (\cos(\theta*\frac{1}{3})+i\sin(\theta*\frac{1}{3})) = 64^{1/3}\cdot (\cos((-\pi/2)*\frac{1}{3})+i\sin((-\pi/2)*\frac{1}{3})$$
$$= 4\cdot (\cos(-\pi/6)+i\sin(-\pi/6))$$
Given that 
$$\cos(-\pi/6)=\frac{\sqrt{3}}{2}$$
and 
$$\sin(-\pi/6) = -\frac{1}{2}$$
We have:
$$4\cdot (\cos(-\pi/6)+i\sin(-\pi/6)) = 4\cdot (\frac{\sqrt{3}}{2}-\frac{1}{2}i) = 2\sqrt{3}-2i$$
The other roots can be found by adding $2\pi$ and $4\pi$ to $\theta$.
So, 
$$4\cdot (\cos((\theta+2\pi)\cdot \frac{1}{3})+i\sin((\theta+2\pi)\cdot \frac{1}{3})) =4i$$
and
$$4\cdot (\cos((\theta+4\pi)\cdot \frac{1}{3})+i\sin((\theta+4\pi)\cdot \frac{1}{3})) = -2\sqrt{3}-2i$$
</s>