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Math Help - complex numbers

  1. #1
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    complex numbers

    i got a few questions...
    how is (e^3x)sin2x=(e^3x)Im(e^2ix)?
    im not very good in complex numbers...

    and then
    to find d56/dt56 [(e^-t)(sint)]
    im confused by
    (-1+i)^56=[ √2 e^(3 πi/4)]^56
    =(2^28)cos42 π + isin42 π
    (-1+i)^56=2^28 <<<<how do you get this??
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  2. #2
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    Quote Originally Posted by yen yen View Post
    i got a few questions...
    how is (e^3x)sin2x=(e^3x)Im(e^2ix)?
    im not very good in complex numbers...

    and then
    to find d56/dt56 [(e^-t)(sint)]
    im confused by
    (-1+i)^56=[ √2 e^(3 πi/4)]^56
    =(2^28)cos42 π + isin42 π
    (-1+i)^56=2^28 <<<<how do you get this??
    The first is a direct application of Euler's formula.

    A complex number can be written in Cartesian co-ordinates:

    z = \mathbf{Re}(z) + i\mathbf{Im}(z)

    or Polar co-ordinates:

    z = \cos{\theta} + i\sin{\theta} = e^{i\theta}.


    Can you see that \cos{\theta} = \mathbf{Re}(z) and \sin{\theta} = \mathbf{Im}(z)?


    Can you see that if \theta = 2x then

    \cos{(2x)} + i\sin{(2x)} = e^{2ix}?


    So \sin{(2x)} is the imaginary part of e^{2ix}.
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  3. #3
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    oh okay... thanks.. now i get it... still need someone to help me with the second problem...
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  4. #4
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    Quote Originally Posted by yen yen View Post
    i got a few questions...
    how is (e^3x)sin2x=(e^3x)Im(e^2ix)?
    im not very good in complex numbers...

    and then
    to find d56/dt56 [(e^-t)(sint)]
    im confused by
    (-1+i)^56=[ √2 e^(3 πi/4)]^56
    =(2^28)cos42 π + isin42 π
    (-1+i)^56=2^28 <<<<how do you get this??
    For part 2, are you asking how you go from

    (-1 + i)^{56} to 2^{28}?


    If so, it's done using DeMoivre's Theorem.


    DeMoivre's Theorem states that:

    If z = r\,\textrm{cis}\,{\theta}

    Then z^n = r^n\,\textrm{cis}\,{(n\theta)} for n \in \mathbf{Z}.


    So to evaluate (-1 + i)^{56}, convert to polars and use DeMoivre's Theorem.


    r = \sqrt{(-1)^2 + 1^2}

     = \sqrt{2}.


    We can see that \cos{\theta} = -1 and \sin{\theta} = 1.

    The function is in the second quadrant.

    So \frac{\sin{\theta}}{\cos{\theta}} = \frac{1}{-1}

    \tan{\theta} = -1

    \theta = \pi - \frac{\pi}{4}

     = \frac{3\pi}{4}.


    Therefore

    -1 + i = \sqrt{2}\,\textrm{cis}\,\left(\frac{3\pi}{4}\right  )


    Now use DeMoivre's Theorem:

    (-1 + i)^{56} = (\sqrt{2})^{56}\,\textrm{cis}\,\left(\frac{56 \cdot 3\pi}{4}\right)

     = 2^{28}\,\textrm{cis}\,(42\pi).


    Now notice that the angle 42\pi is the same as the angle 0, since you have just gone around the unit circle 21 times.


    So (-1 + i)^{56} = 2^{28}\,\textrm{cis}\,0

     = 2^{28}(\cos{0} + i\sin{0})

     = 2^{28}(1 + 0i)

     = 2^{28}.
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