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Math Help - Complex number

  1. #1
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    Complex number

    Hi, I have a cubic equation:

    z^3 + az^2 + bz + c = 0

    where a,b and c are real numbers. I have to show that if z is a solution then so is its complex conjugate, z*.

    It is given that z = -2 -i is a solution to the equation. I must then write down a second complex solution.

    If z = 2 is the thrid solution, what are a,b and c?

    So far I have said that if z = x + iy, then the expansion of the first equation becomes

    (x^3 + 3x^2iy - 3xy^2 - iy^3) + a(x^2 + 2xiy - y^2) + b(x + iy) + c = 0

    when I do the same for the complex conjugate it just gets the x cubed term to become

    (x^3 + 3x^2iy + 3xy^2 + iy^3)

    but that isn't right because the both equations equate to zero and therefore the other, but they are not the same.

    What do I do?

    Any help appreciated.

    (sorry moderators if this is in the wrong topic area)
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  2. #2
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    Quote Originally Posted by Beard View Post
    Hi, I have a cubic equation:

    z^3 + az^2 + bz + c = 0

    where a,b and c are real numbers. I have to show that if z is a solution then so is its complex conjugate, z*.


    This much is true for ANY real polynomial p(x) : its complex non-real roots come as conjugate pairs, because p(\overline{z})=\overline{p(z)} , by the basic properties of the complex conjugation.



    It is given that z = -2 -i is a solution to the equation. I must then write down a second complex solution.

    If z = 2 is the thrid solution, what are a,b and c?


    Then you know that the roots are 2\,,\,-2-i\,,\,-2+i , and the either you compare coefficients of corresponding powers of x in the equality ax^2+bx^2+cx+d=a(x-2)(x-(-2-i))(x-(-2+i)) , or what is the same but more direct and simpler: you remember Viete's formulae

    a_1a_2a_3=-\frac{d}{a}\,,\,\,a_1a_2+a_1a_3+a_2a_3=\frac{c}{a}  \,,\,\,a_1+a_2+a_3=-\frac{b}{a} , with a_1,\,a_2,\,a_3 the polynomial's roots.

    Tonio


    So far I have said that if z = x + iy, then the expansion of the first equation becomes

    (x^3 + 3x^2iy - 3xy^2 - iy^3) + a(x^2 + 2xiy - y^2) + b(x + iy) + c = 0

    when I do the same for the complex conjugate it just gets the x cubed term to become

    (x^3 + 3x^2iy + 3xy^2 + iy^3)

    but that isn't right because the both equations equate to zero and therefore the other, but they are not the same.

    What do I do?

    Any help appreciated.

    (sorry moderators if this is in the wrong topic area)
    .
    Follow Math Help Forum on Facebook and Google+

  3. #3
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    Hi, I have a cubic equation:



    where a,b and c are real numbers. I have to show that if z is a solution then so is its complex conjugate, z*.


    This much is true for ANY real polynomial : its complex non-real roots come as conjugate pairs, because , by the basic properties of the complex conjugation.

    I have to show this though which is the problem because I don't know how.

    (sorry if the answer is really obvious but my maths is poor)
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  4. #4
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    Quote Originally Posted by Beard View Post
    Hi, I have a cubic equation:



    where a,b and c are real numbers. I have to show that if z is a solution then so is its complex conjugate, z*.


    This much is true for ANY real polynomial : its complex non-real roots come as conjugate pairs, because , by the basic properties of the complex conjugation.

    I have to show this though which is the problem because I don't know how.

    (sorry if the answer is really obvious but my maths is poor)

    Check the very basics of complex numbers: this is really easy.

    Tonio
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