Solving cubic roots (with imaginary number)

1. Jul 6, 2009

jeff1evesque

1. The problem statement, all variables and given/known data
(0.1 - 0.3j)^(1/3) = a + bj, where j is the imaginary number or more specifically sqrt(-1).
Does anyone know how to solve for a and b?

2. Relevant equations
I've looked at cubic function equations, along with some polar equations. However, the latter requires some angle to be involved, and I have no such angle. When I try to solve the problem straight out, I get something really crazy.

3. The attempt at a solution
For instance, I get:
(a + bj)^3 = a^3 + 3a^2bi + 3ab^2 - b^3i = 0.1 - 0.3j

thanks,

JL

2. Jul 6, 2009

queenofbabes

You aren't supposed to treat it as a cubic function, that would be crazy.

They're all complex numbers, so what happens when you change them all to polar form?

3. Jul 6, 2009

jeff1evesque

That's the part I would like to know I guess- which I am not sure about how to do. Could you help me, doing it the other way as I expanded the polynomial was just impossible.

4. Jul 6, 2009

queenofbabes

Do you know how to convert to polar form?

What is 0.1 - 0.3j in polar form?
Therefore what is (0.1-0.3j)^1/3 in polar form?

5. Jul 6, 2009

jeff1evesque

to convert to polar form we convert the following:
x = rcos(theta),
y = rsin(theta),
$$z = sqrt(x^2 + y^2)$$

But I can't see how this would convert our equation to polar coordinates. So I guess I don't know how to do the conversion.

6. Jul 6, 2009

queenofbabes

I guess not. I'm referring to the polar form of an imaginary number, expressed as z = r e^(theta)

so for z = a + ib
r = sqrt (a^2 + b^2)
tan (theta) = b/a

It should be in your textbook somewhere....

7. Jul 6, 2009

jeff1evesque

Oh yea, I saw this on wikipedia earlier today, but didn't know how to use it. I've never seen such an equation till today. My multivariate calculus course I don't think covered this.

8. Jul 6, 2009

queenofbabes

This isn't calculus. This is complex numbers....