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Proof of complex exponential

  1. Nov 22, 2004 #1
    [tex]e^{ix}=cosx + isinx [/tex]

    I know this can be easily proven using the Taylor series, but I recall seeing a proof which doesn't use the Taylor series. I'm pretty sure it has something to do with derivatives, but the problem is I don't remember how it went and I can't find it anywhere. So if anyone knows it or has any idea of where to start could you let me know? Thanks.
    Last edited: Nov 22, 2004
  2. jcsd
  3. Nov 22, 2004 #2


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    Once you want to ask that question, you must first ask yourself how you wish to define things like e^z and cos z.
  4. Nov 22, 2004 #3


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    You might consider that if [itex]w = e^{iz}[/itex] then

    [tex]\frac {d^2w}{dz^2} + w = 0[/itex]

    from which

    [tex]w = A \cos z + B \sin z[/tex]

    Does that help?
  5. Nov 23, 2004 #4
    [tex]z=\cos x + i\sin x[/tex]

    [tex]\frac{dz}{dx}=-\sin x + i\cos x=i^2\sin x + i\cos x=i(\cos x + i\sin x )=i.z[/tex]

    [tex]\int\frac{1}{z}\;dz=i\int dx[/tex]


    When x=0, z=1 => c=0.



    [tex]e^{ix}=\cos x + i\sin x[/tex]

    Is that what you're talking about?
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