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Complex Integral

  1. Jun 6, 2004 #1


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    Here is the problem:
    [tex]{\mathop{\rm Im}\nolimits} \int {e^{x(2 + 3i)} } dx[/tex]

    One sec, I'm having another go at it.
    = {\mathop{\rm Im}\nolimits} \int {e^2 } e^{3ix} dx[/tex]
    = {\mathop{\rm Im}\nolimits} \int {e^2 } [\cos (3x) + i\sin (3x)]dx
    = \frac{{ - e^2 \cos (3t)}}{3} \\

    How'd I go?
    Last edited: Jun 6, 2004
  2. jcsd
  3. Jun 6, 2004 #2


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    Not very well, I'm afraid..
    Let us first consider the problem to calculate the antiderivative of the complex exponential,

    This is simply:

    where C is an arbitrary complex constant (I'll set it in the following to 0, for simplicity)

    We are to find the imaginary part:
  4. Jun 6, 2004 #3


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    [tex]e^{x(2+3i)} = e^{2x}e^{3xi} \neq e^{2}e^{3xi}[/tex]

    It's easier than that. Remember that:

    [tex]\int e^{ax} dx = \frac{1}{a}e^{ax} + C[/tex]

    Once you integrate you then need to separate real from imaginary.
    Last edited: Jun 6, 2004
  5. Jun 6, 2004 #4


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    Stuffed up in the first step :(
    Thanks guys. :)
    prolly should goto bed now. :(
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