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Complex Numbers course work

  1. Dec 3, 2009 #1
    1. Find the real part of z=ii by using De Moivre's formula.



    2. Relevant equations
    z= r(cos[tex]\theta[/tex] + i sin[tex]\theta[/tex])
    zn= rn(cos(n[tex]\theta[/tex]) + i sin(n[tex]\theta[/tex]))


    I tried using n=i to solve and got the ans 1i, but somehow feel that its not that simple. And the resultant arguement I got from this approach is i[tex]\theta[/tex] which doesn't make sense. Tried using natural log, but didn't work out too.
     
  2. jcsd
  3. Dec 3, 2009 #2

    Codexus

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    Start by rewriting in exponential form and then use:

    (eix)n = einx

    That should do the trick :wink:
     
  4. Dec 3, 2009 #3
    z = ii = ei(lni)
    so n=lni and the real part is cos(lni). is this correct?
     
  5. Dec 3, 2009 #4

    Codexus

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    I'm not sure where your ln(i) comes from but that part is correct since ln(i) = [tex]i\pi/2[/tex]. However it can be simplified further.

    I would have just written:
    [tex]i^{i} = (e^{i\pi/2})^{i} = e^{i i\pi/2} = e^{- \pi/2}[/tex] and that's your answer since this is a real number already. (Wolfram Alpha confirms it)
     
  6. Dec 3, 2009 #5
    thanks alot. I couldn't get the part where lni = i(PI)/2, tried googling but couldn't find anything. but I could understand your working, you make it look so simple.

    I used the formula, elny = y for my working, thats where the ln comes from. but yours is much more simplified.
     
  7. Dec 3, 2009 #6

    Codexus

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    Well, actually I just used Wolfram Alpha to find that, but if we combine our formulas, we have just proved it's true.
     
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