Finding Principal Root of \sqrt[3]{8i}: Stuck at Arctan?

[MikeH]

I have to find the principal root of $$\sqrt[3]{8 i}$$
But I get stuck at this part
change this to polar coordinates...
$$r= \sqrt {x^2 + y^2}$$
which makes $$r=8$$
but when I try to find $$\theta$$
$$\theta = \arctan \frac{y}{x}$$
from the original x = 0 so how do I find $$\theta$$?

 

[HallsofIvy]

Don't just use formulas without thinking! When you say change "this" to polar coordinates you mean 8i: in the complex plane, that's the point (0, 8)- on the positive imaginary (y) axis which makes a right angle with the real (x) axis- $\theta$ is $\frac{\pi}{2}$ or 90 degrees.

(Of course, $\theta= arctan\frac{y}{x}$ does work even in this case: $tan(\frac{\pi}{2})$ is undefined.}

 

[MikeH]

Thanks for your help, I found the answer