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Math Help - Finding solutions for Complex Numbers

  1. #1
    Junior Member pearlyc's Avatar
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    Finding solutions for Complex Numbers

    Find all solutions, z, of the equation,

    z^4 - z^3 + 27iz - 27i = 0


    expresing your answers in cartesian form.

    ---------------

    use the complex exponential to evaluate  \int e^{2t} sin(3t) dt
    Limits : 0 to pie/3
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pearlyc View Post
    Find all solutions, z, of the equation,

    z^4 - z^3 + 27iz - 27i = 0


    expresing your answers in cartesian form.
    z^4 - z^3 + 27iz - 27i = 0

    z^3(z - 1) - 27i(z - 1) = 0

    (z^3 - 27i)(z - 1) = 0

    (z^3 + (3i)^3)(z - 1) = 0
    should give you a pretty good start.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pearlyc View Post
    use the complex exponential to evaluate  \int e^{2t} sin(3t) dt
    Limits : 0 to pie/3
    I would have thought someone at your level would be able to spell "pi" correctly.

    \int_0^{\pi /3} e^{2t}sin(3t)~dt = Im \left ( \int_0^{\pi / 3}e^{2t}e^{3it}~dt \right )
    where Im( ) stands for the "imaginary part of."

    Can you finish it from here?

    -Dan
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  4. #4
    Junior Member pearlyc's Avatar
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    okay, i got till ..
    Im \left ( \frac {(2-3i)(e^{(2+3i)t}}{13}\right )

    where do i go from here with the limits?
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    Quote Originally Posted by pearlyc View Post
    okay, i got till ..
    Im \left ( \frac {(2-3i)(e^{(2+3i)t}}{13}\right )

    where do i go from here with the limits?
    \text{Im} \, \left ( \frac{2-3i}{13} \left[ (e^{(2+3i)t} \right]_{t = 0}^{t = \pi/3} \right ) = ....
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  6. #6
    Junior Member pearlyc's Avatar
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    I converted exponential part back to the question and I got a zero
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    Quote Originally Posted by pearlyc View Post
    I converted exponential part back to the question and I got a zero
    The answer is not zero. You'll need to show all your working in order for your mistake(s) to be seen.
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  8. #8
    Junior Member pearlyc's Avatar
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     e^{(2+3i)t}

    coverted back to

    e^{2t} sin(3t)

    subbing \frac{\pi}{3} makes sin \pi = 0 subbing it with 0 makes it = 0 too!
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  9. #9
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    Quote Originally Posted by pearlyc View Post
     e^{(2+3i)t}

    coverted back to

    e^{2t} sin(3t)

    subbing \frac{\pi}{3} makes sin \pi = 0 subbing it with 0 makes it = 0 too!
    Is this the sort of work you'd present to your teacher? It's impossible from this 'working' to find your mistakes. Try again - all the working this time please.

    \text{Im} \, \left ( \frac{2-3i}{13} \left[ (e^{(2+3i)t} \right]_{t = 0}^{t = \pi/3} \right ) = ....
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