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Homework Help: Solving cubic roots (with imaginary number)

  1. Jul 6, 2009 #1
    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


  2. jcsd
  3. Jul 6, 2009 #2
    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?
  4. Jul 6, 2009 #3
    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.
  5. Jul 6, 2009 #4
    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?
  6. Jul 6, 2009 #5
    to convert to polar form we convert the following:
    x = rcos(theta),
    y = rsin(theta),
    [tex]z = sqrt(x^2 + y^2)[/tex]

    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.
  7. Jul 6, 2009 #6
    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....
  8. Jul 6, 2009 #7
    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.
  9. Jul 6, 2009 #8
    This isn't calculus. This is complex numbers....
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