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De moivre's theorem complex number

  1. May 31, 2014 #1
    can anyone explain how ro make the working above the red circle to the working in the red circle? why the author do this?
     

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  3. May 31, 2014 #2

    Mentallic

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    It's equivalent to having

    (a+b)c = ac+bc

    where

    [tex]a=z^2+\frac{1}{z^2}[/tex]

    [tex]b=2[/tex]

    [tex]c=z^2-\frac{1}{z^2}[/tex]

    and why he did it should be pretty evident from his next two lines.
     
  4. May 31, 2014 #3

    HallsofIvy

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    It has nothing to do with "complex numbers" or "DeMoivre's Theorem". It is, as mentallic said, just the distributive law.
     
  5. May 31, 2014 #4
    Hahaha, this is something Feynman talked about from his childhood. What you've presented is actually a higher-order form of something called Morrie's[/PLAIN] [Broken] Law (Feynman's little friend in childhood). From what I've studied, a useful application is in the proof of Urquhart's Theorem.
     
    Last edited by a moderator: May 6, 2017
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