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Euler's formula

  1. Feb 7, 2006 #1
    I am to find the imaginary part, real part, square, reciprocal, and absolut value of the complex function:

    [tex]y(x,t)=ie^{i(kx-\omega t)} [/tex]
    [tex]y(x,t)=i\left( cos(kx- \omega t)+ i sin(kx- \omega t) \right)[/tex]
    [tex]y(x,t)=icos(kx- \omega t)-sin(kx- \omega t)[/tex]

    the imaginary part is [tex]cos(kx- \omega t)[/tex]

    the real part is [tex]-sin(kx- \omega t)[/tex]

    the square is:
    [tex]-cos^2(kx- \omega t)-2icos(kx- \omega t)sin(kx- \omega t)+sin^2(kx- \omega t)[/tex]
    [tex]=-cos^2(kx- \omega t)-isin(2kx-2 \omega t)+sin^2(kx- \omega t)[/tex]
    [tex]=-\frac{1}{2}(1+cos(2(kx- \omega t))-isin(2((kx- \omega t))+\frac{1}{2}(1-cos(2(kx- \omega t))[/tex]
    [tex]=-cos(2(kx- \omega t))-isin(2((kx- \omega t))[/tex]

    the reciprocal is:
    [tex]\frac{1}{ie^{i(kx- \omega t)}}[/tex]
    [tex]=-ie^{-i(kx- \omega t)}[/tex]
    [tex]=-icos(kx- \omega t)-sin(kx- \omega t)[/tex]

    absolute value: (not to sure about this...)
    [tex]=|icos(kx- \omega t)-sin(kx- \omega t)|[/tex]
    [tex]cos^2(kx- \omega t)+sin^2(kx- \omega t)[/tex]

    do these look okay?
    Last edited: Feb 7, 2006
  2. jcsd
  3. Feb 8, 2006 #2


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    We usually talk about the norm of a complex number, not its absolute value.

    You're only missing a square root: |x+iy| = [itex]\sqrt{x^2+y^2}[/itex], but it does not affect the answer.

    The rest looks good to me.
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