Solving for Theta in z=(-i)^1/3

[theBEAST]

Homework Statement

First off convert to z=(-i)^1/3.
Here is where I am stuck, how do I convert the above to this form: e^iθ? I am mainly stuck because I don't know how to solve for theta.

I know that θ = arctan(y/x) where y is the coefficient in front of i and x is the real value but in this case there is no x... So you would be dividing by zero which doesn't make sense.

 

[Robert1986]

The "trick" is to write $(-i)^{1/3}$ in the form $re^{i\theta}$ where you must figure out what $r,\theta$ are (this part should be very easy.)

Now you are trying to compute $(re^{i\theta})^{1/3} = r^{1/3}(e^{i\theta})^{1/3}$. Do you see how to do this? Now, there is actually going to be three different solutions, do you see this? Do you see what they are?

 

[scurty]

Alternatively, you can rewrite $z=(-i)^{\frac{1}{3}}$ as $z=e^{log((-i)^\frac{1}{3})}$ and use properties of complex logs to arrive at the correct answer.

 

[vela]

Homework Statement

First off convert to z=(-i)^1/3.
Here is where I am stuck, how do I convert the above to this form: e^iθ? I am mainly stuck because I don't know how to solve for theta.

I know that θ = arctan(y/x) where y is the coefficient in front of i and x is the real value but in this case there is no x... So you would be dividing by zero which doesn't make sense.

Think about where the point z=-i lies in the complex plane. Forget using the formula. What are the polar coordinates of that point?