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Square root of negative complex exponential

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve [itex]\sqrt{-e^{(i2\pi)/3}}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I seem to be missing something simple, as I take:

    [itex]\sqrt{-1}[/itex] = i

    then,

    [itex]e^{(1/2)*(i2\pi)/3}[/itex]

    which comes out as: [itex]ie^{i\pi/3}[/itex]

    however, the solution is:

    [itex]-ie^{i\pi/3}[/itex], and I can't seem to see where that negative is coming from. Any direction would be great, thanks!
     
  2. jcsd
  3. Feb 7, 2012 #2
    The rule sqrt(a*b)=sqrt(a)*sqrt(b) isn't valid for complex numbers.
    Consider sqrt((-1)*(-1))=sqrt(-1)*sqrt(-1)=i*i=(-1) which is incorrect, obviously.
     
  4. Feb 7, 2012 #3

    vela

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    You want to find the principal square root, which is defined to be ##\sqrt{z} = re^{i\theta/2}## when you have ##z = re^{i\theta}## with ##-\pi < \theta \le \pi##. You just need to find the correct ##\theta## for your case.
     
  5. Feb 7, 2012 #4
    So in this case, the correct θ is not [itex] \pi/3 [/itex]? Since, we're trying to solve [itex] \theta/2 [/itex], wouldn't [itex] (2\pi/3)/2 = \pi/3 [/itex] be in our desired range? Also, how would I handle the -1 in this case if that rule is not valid for complex numbers?\

    I think I solved this:

    [itex] -e^{(i2\pi)/3} = e^{-i\pi/3} [/itex], then [itex] \sqrt{e^{-i\pi/3}} = e^{-i\pi/6}[/itex], as desired. This required De Moivre's formula, is there another way to go about this solution?
     
    Last edited: Feb 7, 2012
  6. Feb 7, 2012 #5

    vela

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    $$e^{-i\pi/6} = (-i^2)e^{-i\pi/6} = -i(ie^{-i\pi/6}) = \cdots$$
     
  7. Feb 9, 2012 #6
    The square root of a complex number has two solutions.

    The attachment will help you
     

    Attached Files:

    Last edited: Feb 9, 2012
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