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Math Help - Factorising Imaginary numbers

  1. #1
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    Factorising Imaginary numbers

    I am having trouble finding the poles z1 and z2 of f defined:

    f(z)=\frac{1}{2z^{2}+(6-4i)z-12i}

    Im pretty sure I have to factorise the denominator, but I keep getting the wrong answer which includes imaginary numbers, is it supposed to?
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  2. #2
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    You get 2(z+3)(z-2i) I assume?
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  3. #3
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    That makes sense but how would you factorise when there's an imaginary number involved? Is there a formula of some sort??
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  4. #4
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    Quote Originally Posted by zizou1089 View Post
    That makes sense but how would you factorise when there's an imaginary number involved? Is there a formula of some sort??
    You could use the "-b" formula if you want, i trust your familar enough with that.

    Assuming maddas' solution is correct  f(z) has poles at  2(z+3)(z-2i) = 0 so pole at  z = -3 and  z = 2i

    Yes it is common that in these types of problems that the poles contain imaginary numbers
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  5. #5
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    I tried using the '-b' formula but I get the square root of an imaginary number which I don't think should be the case..
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  6. #6
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    For example how would I factorise:

    (i+\frac{3}{2})z^2-2z+(i-\frac{3}{2})
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