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Homework Help: Getting the argand with De Moires theorem

  1. Mar 16, 2010 #1
    Hi just did an interesting question which made me wonder about solving it another way around

    z = 2 + 5i
    find the value of [/tex]z^{\frac{1}{4}}[/tex] which lies in the second quadrant of the Argand diagram and enter it's argument


    So [tex]arctan(5/2) \approx 1.1902 [/tex]

    We use De Moivre's Thereom and the the oscillation of the trigonometric functions
    [tex]\sqrt{29}(cos(\frac{1.1902 + 2k\pi}{4}) + isin(\frac{1.1902 + 2k\pi}{4}))

    So to lie in the second quadrant the angle must be [tex]\theta > \frac{\pi}{2}[/tex]
    [tex]\theta > \frac{ \pi}{2}[/tex]
    [tex]\frac{1.1902 + 2k\pi}{4} > \frac{\pi}{2}[/tex]

    so k > 0.8 so k > 0 (where k is an integer)

    The next integer after 0 is 1! So sub k = 1 into [tex]\frac{1.1902 + 2\pi}{4}[/tex]

    then convert to degrees 107° agreed???

    BUT how come if we say:
    To lie in the second quadrant [tex]\theta < \pi[/tex]

    [tex]\frac{1.1902 + 2k\pi}{4} < \pi[/tex]
    solves k < 2.8 so if where an integer k can take the ranges 1 to 2.

    BUT if we put 2 in:
    [tex]\frac{1.1902 + 4\pi}{4}[/tex] this is greater than [tex]\pi[/tex]

    Thanks
    Thomas
     
  2. jcsd
  3. Mar 16, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No it doesn't, try solving this inequality again, and show your steps if you get the same result.
     
  4. Mar 16, 2010 #3
    im an idiot :blushing:
     
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