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Imaginary number solution?

  1. Dec 13, 2009 #1
    I'm doing some practice problems for my mechanics exam tomorrow (good ol' SHM) and I can't solve this for the life of me:

    Determine: (-1+i)^(1/3)

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Dec 13, 2009 #2

    nicksauce

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    Write (-1 + i) in polar form (rexp(itheta)). Then multiply the angle by 1/3, and take the cube root of the radius.
     
  4. Dec 13, 2009 #3
    Thanks very much for the answer. IT WORKS!
    However, I don't understnad why/how. Whats is the reasoning behind multiplying then angle and then cube rooting the radius?
     
  5. Dec 13, 2009 #4
    Think about the laws of exponents. If [tex]-1 + i = r e^{ i \theta }[/tex], then [tex](-1 + i)^{ \frac{ 1 }{ 3 } } = ( r e^{ i \theta } )^{ \frac{ 1 }{ 3 } } = \sqrt[3]{r} ( e^{ i \theta } )^{ \frac{ 1 }{ 3 } } = \sqrt[3]{r} e^{ i \frac{ \theta }{ 3 } }[/tex].
     
  6. Dec 13, 2009 #5

    nicksauce

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    Taking the cube root of the radius should be obvious. The radius is a real number, and the cube root satisfies (AB)^1/3 = A^1/3 * B^1/3. Diving the angle by 3 is an application of De Moivre's formula:

    http://en.wikipedia.org/wiki/De_Moivre's_formula
     
  7. Dec 14, 2009 #6
    To multiply two complex numbers in polar form [tex]a \angle b^\circ [/tex] and [tex]c \angle d^\circ[/tex], you multiply the radii (a x c) and add the angles (b + d), so the product is [tex]ac \angle (b+d)^\circ[/tex]

    Cubing a complex number [tex]a \angle b^\circ[/tex] is then [tex](aaa) \angle(b+b+b)^\circ = a^3 \angle(3b)^\circ[/tex]

    So, to find the cube root of a complex number, you can see that you take the cube root of the radius and divide the angle by 3.
     
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