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3rd root of a complex number

  1. Jan 28, 2007 #1
    1. The problem statement, all variables and given/known data
    I need to find the solution to [tex](2-11i)^{\frac{1}{3}}[/tex]


    2. Relevant equations
    If [tex](2-11i)^{\frac{1}{3}}[/tex] were to equal (a + bi) for some real numbers a and b then [tex]2 - 11i = a^3 +3a^2bi-3ab^2-b^3i[/tex]



    3. The attempt at a solution

    From above [tex]a^3-3ab^2 = 2[/tex] and [tex]3a^2b - b^3 = -11[/tex]
    I can factorise (but only slightly) as follows:
    [tex]a(a^2-b^2) = 2 + 2ab^2[/tex]
    [tex]b(a^2-b^2) = -11 - 2a^2b[/tex]

    after losing the a^2-b^2 I'm left with [tex]2b + 2ab^3 = -11a - 2a^3b[/tex] and I can see no other useful factorisations or substitutions :(

    The actual values I need to find here are simple and with not so much guess work found that a = 2 and b = -1. My problem is that I'm not so sure that guess-work is the correct method to be using. I could probably plot both functions and find where there is a point of intersection but is there an algebraic method I can employ?...If so can anyone throw me any pointers?
     
    Last edited: Jan 28, 2007
  2. jcsd
  3. Jan 28, 2007 #2

    mjsd

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    it appears to me that you were trying to express [tex]2-11i[/tex] as a perfect cube [tex](a+ib)^3[/tex] and want to find [tex]a,b[/tex].

    firstly, note that since you are taking a cube root, there will be THREE answers (possibly all complex). The "proper" way to do this is to turn your number [tex]2-11i[/tex] into "polar form" or "exponential form"
    [tex]x+iy \rightarrow r e^{i\theta}[/tex]
    where [tex]r=|x+iy|[/tex] absolute value and [tex]\theta=\text{Arg}(x+iy)[/tex]

    you may now evaluate
    [tex](2-11i)^{1/3} = r^{1/3} e^{i\theta/3}[/tex]

    note you get three answers because
    [tex]\text{Arg}(x+iy) +2n\pi[/tex] is also a valid angle where [tex]n=1,2,\ldots[/tex]
    in this case [tex]\theta/3[/tex] should give you "three" valid angles lie in the range [tex]\left(-\pi,\pi\right][/tex]
     
  4. Jan 28, 2007 #3
    cheers for that mjsd...the book I'm using introduces polar co-ordinates and argand diagrams in the next chapter, (though unfortunately for me, later than the question) I'll return to this question at a later date :smile:
     
  5. Jan 28, 2007 #4
    I think the book is just showing you how much easier these problems become later - the answer to "why do I need to learn a new way to do something I already can do." I love the look on students' faces when I teach them a neat trick that knocks 5 minutes off a 6 minute problem.
     
  6. Jan 29, 2007 #5

    Gib Z

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    I havent done polar forms either, but im assuming that the method you just described employs Eulers Formula?
     
  7. Jan 29, 2007 #6

    mjsd

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    yes: [tex]e^{ix}=\cos x +i \sin x[/tex]
     
  8. Jan 29, 2007 #7

    Gib Z

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    Yup Im familiar with the formula :) Truly a miraculous formula, when I tell people about it they wonder what possible use it could have beyond its beauty, and heres just one example.
     
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