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