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De Moivre's Theorem for Rational Exponents

  1. Dec 18, 2014 #1
    ##cosθ + isinθ = cos(θ + 2kπ) + isin(θ + 2kπ)## for ##k ∈ ℤ##
    ##[cosθ + isinθ]^n = [cos(θ + 2kπ) + isin(θ + 2kπ)]^n##
    ##cos(nθ) + isin(nθ) = cos(nθ + 2nkπ) + isin(nθ + 2nkπ)##
    ##cos(nθ + 2mπ) + isin(nθ + 2mπ) = cos(nθ + 2nkπ) + isin(nθ + 2nkπ)## for ##m ∈ ℤ##

    Now consider the special case ##n = 1/p## for ##p ∈ ℤ##

    ##cos(\frac{θ}{p} + 2mπ) + isin(\frac{θ}{p} + 2mπ) = cos(\frac{θ}{p} + \frac{2kπ}{p}) + isin(\frac{θ}{p} + \frac{2kπ}{p})##

    Is this a contradiction?
     
  2. jcsd
  3. Dec 18, 2014 #2

    mfb

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    Staff: Mentor

    Where do you see a contradiction?
     
  4. Dec 18, 2014 #3

    mathman

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    No. The left side and the right side are two different things. The left side simply expresses the fact [itex]e^{2in\pi}=1[/itex] while the right side simply says there are p pth roots of [itex]e^{i\theta}[/itex]
     
  5. Dec 18, 2014 #4
    The very fact that the RHS and LHS seem to represent two different things is the reason I'm confused as to whether the equality holds.
     
  6. Dec 18, 2014 #5
    Also, could you please elaborate on what the RHS represents?
     
  7. Dec 19, 2014 #6

    mathman

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    [itex]e^{i(\theta+2k\pi)}[/itex] is the same number for all integer values of k. However [itex]e^{\frac{i(\theta+2k\pi)}{p}}[/itex] will have p possible values for k =0,1,...,p-1. These are the pth roots.
     
  8. Dec 19, 2014 #7
    But this means that, technically, they're not equal. Right? They're only equal in special cases where k/p is an integer.
     
  9. Dec 20, 2014 #8

    mathman

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    I hate to keep repeating myself. The left side of the expression represents one particular pth root. The right side is any pth root, depending on k, which is different, unless k/p is an integer.
     
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