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Root of a complex number

  1. Jun 19, 2007 #1
    z^4= 1/2 + i sqrt(3)/2

    I start by transforming into polar form:

    z^4 = e^(i*Pi/3)

    But then I'm blank.
     
  2. jcsd
  3. Jun 19, 2007 #2

    malawi_glenn

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    Have you tried De Moivre's formula ?
     
  4. Jun 19, 2007 #3

    CompuChip

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    I believe the complex root is defined by (that is: it's usually continued from the real function by)
    [tex] \sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2} [/tex]
     
  5. Jun 19, 2007 #4

    malawi_glenn

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    yes, and from De Moivre's formula we get the general:

    [tex] z^{1/n} = r^{1/n}*exp(i \phi / n) [/tex]
     
  6. Jun 19, 2007 #5

    So e^((i Pi)/12) is a solution. How about the other three?
     
  7. Jun 19, 2007 #6

    malawi_glenn

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    Remember that the same complex number can be obtained by rotating 2*pi and so on.

    i.e arg((1/2) + i (rot3 / 2)) = pi / 3 + 2 pi * n, where n is ... -3,-2,-1,0,1,2,3...
     
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