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

  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. D H

    Staff: Mentor

    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. HallsofIvy

    HallsofIvy 40,310
    Staff Emeritus
    Science Advisor

    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. Well in the example I'm doing this is the case.
     
  6. 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. D H

    Staff: Mentor

    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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