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Math Help - Another polar complex number problem

  1. #1
    Newbie Nexus's Avatar
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    Another polar complex number problem

    I've come across another question I'm having difficulty with.

    The fixed points A and B represent the complex numbers a and b in an Argand diagram with origin O.

    By writing a = |a|e^{i\alpha} and b = |b|e^{i\beta}, show that |Im(ab)| = 2\Delta where \Delta is the area of triangle OAB.

    My working takes me through these steps

    ab = |a||b|e^{i\alpha + i\beta}

    ab = |a||b|(cos(\alpha + \beta) +isin(\alpha + \beta))

    Therefore Im(ab) = |a||b|sin(\alpha + \beta)

    However from playing about with diagrams the area of the triangle OAB looks like it should be \Delta = \tfrac{1}{2}|a||b|sin(\alpha - \beta).

    Does anyone have any thoughts on this?
    Last edited by Nexus; September 13th 2010 at 11:06 PM.
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  2. #2
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    You forgot the factor of 1/2 for the area of triangle
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  3. #3
    Newbie Nexus's Avatar
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    Ah so I did. I have corrected that now. What about the sine of the angles though?
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  4. #4
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    Quote Originally Posted by Nexus View Post
    Ah so I did. I have corrected that now. What about the sine of the angles though?
    I assume that A is located in first quadrant and B is located in the fourth quadrant, then the area of the triangle will be 1/2 |a||b|sin (α + β).

    Are there any restrictions to the angles? Because I think this only works if we consider case like mine or similar.
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  5. #5
    Newbie Nexus's Avatar
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    No other restrictions were stated in the question. I think that if the question were modified to say b = |b|e^{-\beta} then this does lead to |Im(ab)| = |a||b||sin(\alpha - \beta)| = 2\Delta.

    The book I am using is Pure Mathematics 4 by Hugh Neill and Douglas Quadling and although it is nice and accessible I have found a fair few questions which I have had to assume contain errors. I think I am going to have to add this one to that list.

    Thanks for the input though!
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