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Optics: Complex exponentials for sine

  1. Jan 31, 2013 #1
    1. The problem statement, all variables and given/known data
    Imagine that we have two waves of the same amplitude, speed and frequency over-lapping in some region of space such that the resultant disturbance is

    [itex]\psi(y,t) = Acos(ky+\omega t) + Acos(ky-\omega t +\pi)[/itex]

    Using complex exponentials show that


    [itex]\psi(y,t) = -2Asin(ky) sin(\omega t)[/itex]


    2. Relevant equations

    [itex]Ae^{i\theta} = Acos(\theta) + Aisin(\theta) [/itex]

    3. The attempt at a solution

    I actually have the solution worked out, but my issue occurred once consulting the solutions manual. I ended up with:

    [itex]\psi(y,t) = A(2icos(ky)sin(\omega t)-2sin(ky) sin(\omega t))[/itex]

    with the real part being the solution:

    [itex]\psi(y,t) = -2Asin(ky) sin(\omega t)[/itex]



    Then the solutions manual says "Had we begun with sine waves, i.e.

    [itex]\psi(y,t) = Asin(ky+\omega t) + Asin(ky-\omega t +\pi)[/itex]

    the treatment would have been identical up until the last step, where, this time, the imaginary part would be taken to give:"

    [itex]\psi(y,t) = 2Acos(ky)sin(\omega t)[/itex]


    I'm confused as to why we can just choose the imaginary part because we happen to have started with sine functions. I thought the real part was always used and the imaginary part ignored. I think I'm missing some intuition with converting sine functions into complex exponentials... I understand that

    [itex]Acos(\theta) = Ae^{i\theta}[/itex] because we can ignore the [itex]Aisin(\theta)[/itex] term,
    but what if we start with [itex]Asin(\theta)[/itex]? How is this changed to an exponential?
     
  2. jcsd
  3. Jan 31, 2013 #2

    Simon Bridge

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    You don't. The cosine-sine part is the real part when you start with sine functions.
    This is because:

    ##\sin\theta = \frac{1}{i}\left ( e^{i\theta}-e^{-i\theta} \right )##

    ----------------

    LaTeX note: put a backslash in front of function names to get them to typeset properly ... sin\theta gives you ##sin\theta## while \sin\theta gives you ##\sin\theta## ...
     
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