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Euler's identity

  1. Aug 13, 2008 #1
    How does y = e^(x(1-i)) + e^(x(1+i))
    work out to y = (e^x)sinx + (e^x)cosx???

    Using Euler's identity I get,

    y = (e^x)e^-ix + (e^x)e^ix
    y = e^x(cosx - isinx + cosx + isinx)
    y = e^x(2cosx)
  2. jcsd
  3. Aug 13, 2008 #2
    why don't you write your equations using [tex] tabs...
    it will be more readable by others...
  4. Aug 13, 2008 #3


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    It doesn't.
  5. Aug 13, 2008 #4


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    Obviously, by Euler's identity, (e^x)sinx+ (e^x)cosx= e^(x(1+i)) only, not that sum.
  6. Aug 13, 2008 #5


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    Halls meant (e^x)*i*sinx+ (e^x)cosx. The expansion you did, 2cos(x)e^x, is correct. and it's not equal to sin(x)e^x+cos(x)e^x.
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