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Math Help - Factorisation

  1. #1
    Junior Member phgao's Avatar
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    Factorisation

    This isn't urgent, but I'd like explanations + working out so I can understand the question. THanks!

    Factorise:

    2z^2 + 2(z-2)z +3/2 - 6i
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by phgao
    This isn't urgent, but I'd like explanations + working out so I can understand the question. THanks!

    Factorise:

    2z^2 + 2(z-2)z +3/2 - 6i
    This is a complex quadratic, so you can proceed by completing the square,
    or using the quadratic formula to find its roots a and b then the
    factorisation is:

    (z-a)(z-b).

    So lets simplify the equation a bit:

    2z^2 + 2(z-2)z +3/2 - 6\bold{i}=4z^2-4z+(3/2 - 6\bold{i})

    The only problem will be finding the square root of the discriminant.
    In this case the discriminant is:

    16(6 \bold{i}-1/2)=8(12\bold{i}-1)

    We need the square root of this. We proceed by converting the discriminant
    to polar form, then finding the square root by square-rooting the modulus and
    halving the argument. Or as MathWorld would put it:

    \sqrt{x+iy}= \pm(x^2+y^2)^{1/4}(\cos(1/2\  \mbox{atan} (y/x))+\bold{i} \sin(1/2\  \mbox{atan} (y/x))),

    Which should allow you to complete the factorisation.

    RonL
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