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Complex Solutions

  1. Jan 10, 2009 #1
    1. Hi I posted a previous question on this forum and it was answered well but I had another question about complex solutions/roots.

    for example if i have a question like

    z^2 = 1e^(j)(pie)

    z = 1e^(j)(pie + 2kpie)^1/2 k =0,1

    1. z = 1e^(j)(pie/2 ) = 0 + j
    2. z = 1e^(j)(3pie/2) = 0 - j

    if I test these solutions (0+j)(0+j) = -1 , (0 - j)(0 - j) = -1

    they are correct but my question is are those angles correct????

    another way to solve that question seems to be

    z^2 = 1e^(j)(-pie)

    z = 1e^(j)(-pie + 2kpie)^1/2 k =0,1

    1. z = 1e^(j)(-pie/2 ) = 0 - j
    2. z = 1e^(j)(pie/2) = 0 + j


    if i go backwards

    0 + j = 1e^(j)(-pie/2 ) and 0 - j = 1e^(j)(pie/2)


    so which angles are correct???

    Like on an exam I don`t really know which angles I would write!
     
  2. jcsd
  3. Jan 10, 2009 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Both angles are correct; the complex exponential is periodic with period 2pi, so if you add 2pi to your angle of -pi/2, you get the same value for the exponential

    i.e. [tex]e^{j\frac{-\pi}{2}}=e^{j\left(\frac{-\pi}{2}+2\pi\right)}=e^{j\frac{3\pi}{2}}=-j[/tex]
     
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