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Trigonometric Complex Numbers

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    how are the solutions of the fourth roots pi/2? how do you get pi/2, you know the thing after "cos" and "isin" ?
     
  2. jcsd
  3. Apr 14, 2008 #2
    First write it in polar form, or trigonometric form, however you call it, and after that use de moivre's formula to find its roots.

    let z be a complex nr.

    z=a+bi, writing it in polar forms : [tex]a=r cos(\theta),b=\ro sin\theta[/tex]


    So,

    [tex] z=r (cos\theta+isin\theta)[/tex]

    now

    [tex]z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos\frac{\theta +2k\pi}{n}+isin{\frac{\theta+2k\pi}{n})[/tex]

    Now all you need to do is figure out what [tex]\theta [/tex] is, and your fine.

    Or if you want the exponential representation of a complex nr:

    [tex]e^{ix}=cosx+isinx[/tex]

    [tex]e^{i\frac{\pi}{2}}=i[/tex] so we get

    [tex] i=cos\frac{\pi}{2}+isin\frac{\pi}{2}[/tex]
     
    Last edited: Apr 14, 2008
  4. Apr 14, 2008 #3
    how do you know that x = pi/2
     
  5. Apr 14, 2008 #4
    well z=i, is a complex nr right. Following my elaboration above we have a=0, b=1, right?

    so [tex]\theta =\arctan\frac{1}{0}-->\frac{\pi}{2}[/tex] loosly speaking.

    so the exponential form of i is what i wrote [tex] i=e^{i\frac{\pi}{2}}[/tex]

    SO to find the four roots of i just follow de moivres formula that i wrote above, letting k=0,1,2,3.
     
  6. Apr 14, 2008 #5
    how do you know a=0 and b=1? isnt it a=1 and b=-1 or did i do something wrong
     
  7. Apr 15, 2008 #6

    HallsofIvy

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    Your number is i. If you write that in the form a+ bi, what are a and b?
     
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