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Complex Integral

  1. Jun 6, 2004 #1

    dcl

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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.
    [tex]
    = {\mathop{\rm Im}\nolimits} \int {e^2 } e^{3ix} dx[/tex]
    [tex]
    = {\mathop{\rm Im}\nolimits} \int {e^2 } [\cos (3x) + i\sin (3x)]dx
    [/tex]
    [tex]
    \begin{array}{l}
    = \frac{{ - e^2 \cos (3t)}}{3} \\
    \end{array}
    [/tex]

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

    arildno

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    Not very well, I'm afraid..
    Let us first consider the problem to calculate the antiderivative of the complex exponential,
    [tex]\int{e}^{(2+3i)x)}dx[/tex]

    This is simply:
    [tex]\int{e}^{(2+3i)x)}dx=\frac{1}{2+3i}{e}^{(2+3i)x)}+C[/tex]

    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:
    [tex]Im(\frac{1}{2+3i}{e}^{(2+3i)x)})=Im(\frac{2-3i}{13}{e}^{(2+3i)x)})[/tex]
    or:
    [tex]Im(\frac{1}{2+3i}{e}^{(2+3i)x)})=\frac{e^{2x}}{13}(2\sin(3x)-3\cos(3x))[/tex]
     
  4. Jun 6, 2004 #3

    Zurtex

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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

    dcl

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