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Math Help - Polynomial with complex roots

  1. #1
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    Polynomial with complex roots

    Hey guys I'm not sure whether my approach to this question is right...kinda looks simple enough eh?

    QUESTION
    One root of  x^{2}+(1-i)x+K=0 is 2+3i. Find the other root of this equation and the value of K in the form a+ib



    So this is what I did:

    Since 2+3i is a root, 2-3i is also a root as complex roots come in conjugate pairs

    (x-(2+3i))(x-(2-3i))=x^{2}-4x+13

    But then?
    \Sigma\alpha=\alpha+\beta=(2-3i)+(2+3i)=\frac{-b}{a}=4\neq1-i

    have I misunderstood something?...im rusty with complex numbers

    The actual answers are that the other root is -3-2i,and K=-13i

    Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by aonin View Post
    Since 2+3i is a root, 2-3i is also a root as complex roots come in conjugate pairs

    That is true if the polynomial has real coefficientes, an this is not the case. Substitute x=2+3i and you will find K.


    Fernando Revilla
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  3. #3
    MHF Contributor Swlabr's Avatar
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    If a polynomial has REAL coefficients, then what you said was correct. That is, roots come in complex conjugate pairs. However, this polynomial has a complex coefficient, so what you said doesn't hold (sorry!).

    Personally, I would just work out (x-2-3i)(x-a-bi), using the fact that the coefficient of x must be (1-i). It is, perhaps, crude. But it works.
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  4. #4
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    Quote Originally Posted by aonin View Post
    Hey guys I'm not sure whether my approach to this question is right...kinda looks simple enough eh?

    QUESTION
    One root of  x^{2}+(1-i)x+K=0 is 2+3i. Find the other root of this equation and the value of K in the form a+ib



    So this is what I did:

    Since 2+3i is a root, 2-3i is also a root as complex roots come in conjugate pairs

    (x-(2+3i))(x-(2-3i))=x^{2}-4x+13

    But then?
    \Sigma\alpha=\alpha+\beta=(2-3i)+(2+3i)=\frac{-b}{a}=4\neq1-i

    have I misunderstood something?...im rusty with complex numbers

    The actual answers are that the other root is -3-2i,and K=-13i

    Thanks

    Put 2+3i in the equation and find the value of k -> k=-3i

    then put the k in the equation and find two X (quadratic function)
    note that sqrt(50i)=5+5i
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  5. #5
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    Quote Originally Posted by ahaok View Post
    Put 2+3i in the equation and find the value of k -> k=-3i

    then put the k in the equation and find two X (quadratic function)
    note that sqrt(50i)=5+5i
    Thanks!
    Last edited by aonin; January 19th 2011 at 03:28 AM. Reason: lol..careless mistake
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  6. #6
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    Quote Originally Posted by aonin View Post
    Hey, I got this..and I've not learnt how to obtain the roots so could you show me the working out after this step pls

    k=-13i

    x^{2}+(1-i)x-13i=0

    Im not sure how to find the roots? uisng square roots of complex numbers lol? let z^{2}=x^{2}-y^{2}+2xyi?
    infact you must ontain the roots of a quadratic function:
    Quadratic function - Wikipedia, the free encyclopedia
    in this case:
    a=1, b=1-i, c=-13i

    X1,2=(-b+-sqrt(b^2-4ac))/2a
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ahaok View Post
    infact you must ontain the roots of a quadratic function:
    Quadratic function - Wikipedia, the free encyclopedia
    in this case:
    a=1, b=1-i, c=-13i

    X1,2=(-b+-sqrt(b^2-4ac))/2a
    This is all a bit too much though - as the OP pointed out, to do it this way you need to find the square root of a complex number, and so we are spiralling down a route that, although correct, requires a lot of work.

    It is much easier to approach the question in the way I pointed out in my post.

    (x-2-3i)(x-a-bi) = x^2 + (1-i)x + K
    \Rightarrow x^2 - (a+2+(b+3)i)x +(2+3i)(a+ib) = x^2 + (1-i)x + K
    \Rightarrow -(a+2+(b+3)i) = 1-i

    So, what are a and b? So what is the other root, and what is K?

    Simples...
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