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Math Help - Sin of Complex Number

  1. #1
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    Sin of Complex Number

    Express in the form of x + jy:
    sin(\frac{5\pi}{6} + j)

    I know i is the more common way of representing the imaginary number, but I'm an engineer so they say I should get used to j as to not conflict with something else (I can't remember what...)

    Anyway, my attempt.
    \frac{e^{j({\frac{5\pi}{6} + j})}-e^{-j({\frac{5\pi}{6} + j})}}{2j}
    \frac{e^{-1+j\frac{5\pi}{6}}-e^{1-j\frac{5\pi}{6}}}{2j}
    \frac{e^{-1}(cos(\frac{5\pi}{6})+jsin(\frac{5\pi}{6}))-e(cos(-\frac{5\pi}{6})+jsin(-\frac{5\pi}{6}))}{2j}

    And now I have no idea what to do. Am I on the right track, even?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by BlackBlaze View Post
    Express in the form of x + jy:
    sin(\frac{5\pi}{6} + j)

    I know i is the more common way of representing the imaginary number, but I'm an engineer so they say I should get used to j as to not conflict with something else (I can't remember what...)

    Anyway, my attempt.
    \frac{e^{j({\frac{5\pi}{6} + j})}-e^{-j({\frac{5\pi}{6} + j})}}{2j}
    \frac{e^{-1+j\frac{5\pi}{6}}-e^{1-j\frac{5\pi}{6}}}{2j}
    \frac{e^{-1}(cos(\frac{5\pi}{6})+jsin(\frac{5\pi}{6}))-e(cos(-\frac{5\pi}{6})+jsin(-\frac{5\pi}{6}))}{2j}

    And now I have no idea what to do. Am I on the right track, even?
    You are doing perfect so far.

    All you need to do now is simplify the trig expressions and group the real and imaginary parts.

    I'll start you off. Your last like of work is the same as:

    \frac{-\frac{\sqrt{3}}{2}e^{-1}+\frac{1}{2}e^{-1}j+\frac{\sqrt{3}}{2}e-\frac{1}{2}ej}{2j}=\dots

    Can you continue? Remember to multiply both numerator and denominator by \frac{-j}{-j} when you finish simplifying (to avoid having complex numbers in the denominator of our final results).
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  3. #3
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    Quote Originally Posted by BlackBlaze View Post
    Express in the form of x + jy:
    sin(\frac{5\pi}{6} + j)
    It really helps to know the basic elementy functions.
    \sin(x+yj)=\sin(x)\cosh(y)+j\,\cos(x)\sinh(y).
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  4. #4
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    @Plato, I'm not exactly sure if I'm allowed to do that. I can't seem to find that in my textbook...it'd certainly be a lot easier than doing this whole exponential thing though. I'll keep it in mind, so thank you for that.

    @Chris, can you confirm my answer? I got:
    \frac{1}{4}(e^{-1}+e)+j(\frac{\sqrt{3}}{2}(e^{-1}-e))
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