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Math Help - Complex numbers with a polenom!

  1. #1
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    Question Complex numbers with a polenom!

    Hello,

    So here is this combined polynomial & complex numbers; I couldn't solve it!

    F(z)= z^7 + a(6)*z^6 + a(5)*z^5 + a(4)*z^4 + a(3)*z^3 + a(2)*z^2 + a(1)*z^1+a(0)

    It is given that z=-4-2i is a root of F(z),F'(z),F''(z) (I guess it's correct to say that z=-4+2i is also root for the three functions!)

    a(0)=-6

    a(0),,,,,,a(6) are all real numbers

    What is the value of a(6) ?

    **a(Number) indicates for index!

    Any thoughts?
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  2. #2
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    Here's one thought: if the coefficients of the polynomial are all real, then any complex roots come in complex-conjugate pairs. Hence, -4+2i is also a root of F, F', and F''. That means you technically have 7 equations and 7 unknowns (the unknowns being the coefficients). I would just start plugging in numbers into the functions you have, and start cranking away. It's a fair bit of algebra, looks like. What do you get?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    Here's one thought: if the coefficients of the polynomial are all real, then any complex roots come in complex-conjugate pairs. Hence, -4+2i is also a root of F, F', and F''. That means you technically have 7 equations and 7 unknowns (the unknowns being the coefficients). I would just start plugging in numbers into the functions you have, and start cranking away. It's a fair bit of algebra, looks like. What do you get?
    Are you sure this is a good was to solve this it mayet take me hours
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  4. #4
    A Plied Mathematician
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    Are you allowed to use a Computer Algebra System like Mathematica?
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  5. #5
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    Since -4-2i is a zero of F, F', and F'', (z + 4 + 2i)^3 must be a factor of F. The same is true of -4+2i, so (z + 4 -2i)^3 is also a factor. Let's say the remaining zero of F is a. Then

    F(z) = (z-a) (z + 4 + 2i)^3 (z + 4 -2i)^3
    = (z-a) [(z + 4 + 2i) (z + 4 -2i)]^3
    = (z-a) (z^2 + 8z + 20)^3

    You can use F(0) = -6 to determine a. So...
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  6. #6
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    Quote Originally Posted by awkward View Post
    Since -4-2i is a zero of F, F', and F'', (z + 4 + 2i)^3 must be a factor of F. The same is true of -4+2i, so (z + 4 -2i)^3 is also a factor. Let's say the remaining zero of F is a. Then

    F(z) = (z-a) (z + 4 + 2i)^3 (z + 4 -2i)^3
    = (z-a) [(z + 4 + 2i) (z + 4 -2i)]^3
    = (z-a) (z^2 + 8z + 20)^3

    You can use F(0) = -6 to determine a. So...
    Very nice! Much more elegant than my brute force solution. There's your answer, kadmany.
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  7. #7
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    Hello,first thank u all,

    second off "Computer Algebra System" please if u got a link to such a calculator then post it,

    second the method with

    F(z) = (z-a) (z + 4 + 2i)^3 (z + 4 -2i)^3
    = (z-a) [(z + 4 + 2i) (z + 4 -2i)]^3
    = (z-a) (z^2 + 8z + 20)^3

    You can use F(0) = -6 to determine a. So...

    I put z=0 then I get (-a)[(4+2i)(4-2i)]^3 = (-a)(20)^3

    this equals to -6 => a*20^3=6 => a=6/(20^3)

    then what?, I gotta find a(6) to open operators is bit hard!
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  8. #8
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    Guys I got this expression :



    Is it correct?
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