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Math Help - Multiplying imaginary factors of a polynomial

  1. #1
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    [Solved] Multiplying imaginary factors of a polynomial

    The problem:
    (x - e^{\frac{i\pi}{4}})(x - e^{\frac{-i\pi}{4}})

    My attempt:
    I used the distributive rule to expand it, and got:

    x^2 - x(e^{\frac{i\pi}{4}} + e^{\frac{-i\pi}{4}}) + 1

    I recognise that cos(\frac{\pi}{4}) = \frac{1}{2}(e^{\frac{i\pi}{4}} + e^{\frac{-i\pi}{4}})

    So I get x^2 - x(2cos(\frac{\pi}{4})) + 1
    = x^2 - \frac{2x}{\sqrt{2}} + 1
    = x^2 - \sqrt{2}x + 1

    I'm unsure if this is correct, since I don't have a solution. Any assistance would be greatly appreciated.
    Last edited by Glitch; August 20th 2010 at 08:19 AM. Reason: I found a mistake, and have corrected it.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The problem:
    (x - e^{\frac{\pi}{4}})(x - e^{\frac{-\pi}{4}})

    My attempt:
    I used the distributive rule to expand it, and got:

    x^2 - x(e^{\frac{\pi}{4}} + e^{\frac{-\pi}{4}}) + 1

    I recognise that cos(\frac{\pi}{4}) = \frac{1}{2}(e^{\frac{\pi}{4}} + e^{\frac{-\pi}{4}})

    So I get x^2 - x(2cos(\frac{\pi}{4})) + 1
    = x^2 - \frac{2x}{\sqrt{2}} + 1
    = x^2 - \sqrt{2}x + 1

    I'm unsure if this is correct, since I don't have a solution. Any assistance would be greatly appreciated.
    \cos x = \frac12(e^{ix} + e^{-ix}) (notice the i's in the exponents). So either the problem should read (x - e^{\frac{\pi i}{4}})(x - e^{\frac{-\pi i}{4}}), or the solution should have cosh instead of cos. If the problem should have had i's in the exponents then the solution is correct.
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  3. #3
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    Whoops, yeah, there's supposed to be 'i'. Thanks.
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