Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook