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De Moivre's Theorum

  1. Apr 25, 2007 #1
    De Moivre's Theorum - (just needs checking)

    1. The problem statement, all variables and given/known data

    (1+i)^20

    2. Relevant equations

    De Moivre's theorum: [r(cos theta + isin theta)]^n= r^n(cos ntheta + isin ntheta):uhh: (i think)

    3. The attempt at a solution
    x= 1
    y=1
    r= 1
    theta= 45 degrees
    [1(cos 45 + i sin 45)]^20 = 1^20[cos(20*45) + i sin (20*45)]
    therefore:
    1(cos 900 + i sin 900)
    (-1 + i 0)
    the answer: is -1 ??:biggrin:
     
    Last edited: Apr 25, 2007
  2. jcsd
  3. Apr 25, 2007 #2
    Yes looks like it.
     
  4. Apr 25, 2007 #3
    are you sure?
     
  5. Apr 25, 2007 #4

    George Jones

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    Check out da_willem's comment.

    De Moivre's theorem is the standard way to do this. Another way is

    [tex]\left( 1 + i \right)^{20} = \left( \left( \left( 1 + i \right)^2 \right)^2 \right)^5.[/tex]

    Work from the inside to the outside.
     
  6. Apr 25, 2007 #5

    HallsofIvy

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    If z= 1+ i, then [itex]r= |z|= \sqrt{(1+i)(1-i)}= \sqrt{2}[/itex], NOT 1. Your twentieth power is missing a factor of [itex]r^{20}= (\sqrt{2})^20= 2^{10}= 1024[/itex].
     
  7. Apr 26, 2007 #6
    or is it the square root of 2?
     
  8. Apr 26, 2007 #7

    Dick

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    How would you figure out the answer to that question?
     
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