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.
 A: Let  $y=(-64i)^{\frac13}\implies y^3=-64i=64i^3=(4i)^3$
So,$y^3-(4i)^3=0$
$(y-4i)\{y^2+y\cdot 4i+(4i)^2\}=0$
If $y-4i=0,y=4i$
else $y^2+y\cdot 4i-16=0\implies y=\frac{-4i\pm\sqrt{(-4i)^2-4\cdot1(-16)}}2=\pm2\sqrt3-2i$
So, $y=4i,\pm2\sqrt3-2i$ 
A: 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
$$4\cdot (\cos((\theta+4\pi)\cdot \frac{1}{3})+i\sin((\theta+4\pi)\cdot \frac{1}{3})) = -2\sqrt{3}-2i$$
A: For any $n\in\mathbb{Z}$,
$$\left(-64i\right)^{\frac{1}{3}}=\left(64\exp\left[\left(\frac{3\pi}{2}+2\pi n\right)i\right]\right)^{\frac{1}{3}}=4\exp\left[\left(\frac{\pi}{2}+\frac{2\pi n}{3}\right)i\right]=4\exp\left[\frac{3\pi+4\pi n}{6}i\right]=4\exp \left[\frac{\left(3+4n\right)\pi}{6}i\right]$$
The cube roots in polar form are:
$$4\exp\left[\frac{\pi}{2}i\right]
\quad\text{or}\quad
4\exp\left[\frac{7\pi}{6}i\right]
\quad\text{or}\quad
4\exp\left[\frac{11\pi}{6}i\right]$$
and in Cartesian form:
$$4i
\quad\text{or}\quad
-2\sqrt{3}-2i
\quad\text{or}\quad
2\sqrt{3}-2i$$
