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Math Help - complex numbers

  1. #1
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    complex numbers

    Express in the form a + bi, where a and b are real

    (a) sin(3pi/2 - i)

    (b) cosh(1 + i)

    have tried using addition formula on (a) but its not right.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ardam View Post
    Express in the form a + bi, where a and b are real

    (a) sin(3pi/2 - i)

    (b) cosh(1 + i)

    have tried using addition formula on (a) but its not right.
    Note that for z\in\mathbb{C}, \sin(z)=\frac{e^{iz}-e^{-iz}}{2i}

    Thus, \sin\left(\tfrac{3\pi}{2}-i\right)=\frac{e^{\frac{3\pi}{2}i+1}-e^{-\frac{3\pi}{2}i-1}}{2i} =\frac{e\left(\cos\left(\tfrac{3\pi}{2}\right)+i\s  in\left(\tfrac{3\pi}{2}\right)\right)-e^{-1}\left(\cos\left(-\tfrac{3\pi}{2}\right)+i\sin\left(-\tfrac{3\pi}{2}\right)\right)}{2i}=\frac{-ei-e^{-1}i}{2i} =-\frac{e+e^{-1}}{2}=\color{red}\boxed{-\cosh(1)}

    Does this make sense?

    Try part (b). Note that if z\in\mathbb{C}, then \cosh(z)=\frac{e^z+e^{-z}}{2}.
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Note that for z\in\mathbb{C}, \sin(z)=\frac{e^{iz}-e^{-iz}}{2i}

    Thus, \sin\left(\tfrac{3\pi}{2}-i\right)=\frac{e^{\frac{3\pi}{2}i+1}-e^{-\frac{3\pi}{2}i-1}}{2i} =\frac{e\left(\cos\left(\tfrac{3\pi}{2}\right)+i\s  in\left(\tfrac{3\pi}{2}\right)\right)-e^{-1}\left(\cos\left(-\tfrac{3\pi}{2}\right)+i\sin\left(-\tfrac{3\pi}{2}\right)\right)}{2i}=\frac{-ei-e^{-1}i}{2i} =-\frac{e+e^{-1}}{2}=\color{red}\boxed{-\cosh(1)}

    Does this make sense?

    Try part (b). Note that if z\in\mathbb{C}, then \cosh(z)=\frac{e^z+e^{-z}}{2}.
    cheers for that.
    right so cosh(z)

    well sub 1 + i in

    \cosh(z)=\frac{e^{1+i} +e^{-1-i}}{2}

    then can you take e outside bracket or treat e^1+i as e^i.e?
    Last edited by ardam; January 28th 2009 at 04:10 PM.
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  4. #4
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    anyone please?
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  5. #5
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    Quote Originally Posted by ardam View Post
    anyone please?
    Why don't you know anything about elementry functions?
    LIKE:
    \sin (z) = \sin (x)\cosh (y) + i\cos (x)\sinh (y)

    \cos (z) = \cos (x)\cosh (y) - i\sin (x)\sinh (y)

    \sinh (z) = \sinh (x)\cos (y) + i\cosh (x)\sin (y)

    \cosh (z) = \cosh (x)\cos (y) - i\sinh (x)\sin (y)

    These are standard and easy to derive.
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