I'm having (another) thick moment (complex numbers)

  1. Nov 19, 2006 #1
    This is a question thats stumping both myself, and my friends who are on maths degrees!

    So...

    cos(x) can be written as [tex]\frac{1}{2}(e^{ix}+e^{-ix})[/tex] correct?

    so does that make its conjugate [tex]\frac{1}{2}(e^{-ix}+e^{ix})[/tex], i.e. cos(x) again? or does the switching of the sign go in front of the e? Its been a long time since I used complex numbers, so I (and my friends) are a little rusty! Any help would be appreciated.

    Thanks

    Brewer
     
  2. jcsd
  3. Nov 19, 2006 #2

    D H

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    You are assuming that [tex]e^{iz}[/tex] and [tex]e^{-iz}[/tex] are conjugates of each other. This is only true if z is pure real.

    If [tex]z=x+iy, (x,y)\in\mathbb R\times \mathbb R)[/tex], then [tex]e^{iz}=e^{-y}e^{ix}[/tex] and [tex]e^{-iz}=e^{y}e^{-ix}[/tex]. Taking the conjugates, [tex]e^{iz^\ast}=e^{-y}e^{-ix} \ne e^{-iz}[/tex] and [tex]e^{-iz^\ast}=e^{y}e^{ix} \ne e^{iz}[/tex].
     
    Last edited: Nov 19, 2006
  4. Nov 19, 2006 #3

    HallsofIvy

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    In order that your equation [tex]\frac{1}{2}(e^{ix}+e^{-ix})[/tex] be correct, x must be a real number and then cos(x) is a real number. Is the complex conjugate of cos(x) equal to cos(x)? Of course it is: the complex conjugate of any real number is itself!
     
  5. Nov 19, 2006 #4
    Well in the example I'm doing this is the case.
     
  6. Nov 19, 2006 #5
    Good that makes me feel better, as thats the reasoning I came up with, and the other thought was conceived by 2 maths students!
     
  7. Nov 19, 2006 #6

    D H

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    Halls, you should know better!

    The equation [tex]\cos(x) = \frac{1}{2}(e^{ix}+e^{-ix})[/tex] follows directly from Euler's formula, [tex]e^{ix} = \cos(x) + i\sin(x)[/tex], which is valid for all real and complex x. Thus the given expression for [tex]\cos(x)[/tex] is valid for all real and complex x.
     
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