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Math Help - Complex roots of a quadratic equation

  1. #1
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    Complex roots of a quadratic equation

    The question:

    Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation ax^2 + bx + c = 0 with a, b, c being real coefficients, then so is the conjugate of 'z'.

    My attempt:

    Let z = z_{1} \pm z_2i

    \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = z_{1} \pm z_2i

    \frac{1}{2a}(-b \pm \sqrt{b^2 - 4ac}) = z_{1} \pm z_2i

    Basically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.

    Thanks!
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  2. #2
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    The roots of ax^2 + bx + c = 0 are

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.


    If b^2 - 4ac < 0 then that means \sqrt{b^2 - 4ac} is imaginary. So we can say \sqrt{b^2 - 4ac} = ni, where n is some multiple of i.


    So x = \frac{-b \pm n i}{2a}

     = -\frac{b}{2a} \pm \left(\frac{n}{2a}\right)i.


    So the two roots are complex conjugates.
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    Ahh, thanks. So would my solution be acceptable, or should I simplify it like yours?
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    Quote Originally Posted by Glitch View Post
    Ahh, thanks. So would my solution be acceptable, or should I simplify it like yours?
    To be honest, I can't really follow your solution...
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    It's pretty much the same as yours, except you replaced the square root with 'ni'. I was trying to show that the general form of a complex number is equivalent to a simplified quadratic formula.
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    Quote Originally Posted by Glitch View Post
    The question:
    Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation ax^2 + bx + c = 0 with a, b, c being real coefficients, then so is the conjugate of 'z'.
    The standard way of proving the is to use properties of the complex conjugate.
    If a is a real number then \overline{a}=a.
    The conjugate of a sum is the sum of conjugates.
    Therefore we have:

    \overline {az^2  + bz + c}  = \overline 0  = 0\, \Rightarrow \,a\overline z ^2  + b\overline z  + c = 0
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    Quote Originally Posted by Glitch View Post
    The question:

    Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation ax^2 + bx + c = 0 with a, b, c being real coefficients, then so is the conjugate of 'z'.

    My attempt:

    Let z = z_{1} \pm z_2i

    \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = z_{1} \pm z_2i

    \frac{1}{2a}(-b \pm \sqrt{b^2 - 4ac}) = z_{1} \pm z_2i
    But at this point you haven't yet said that \sqrt{b^2- 4ac} is an imaginary number!

    Basically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.

    Thanks!
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  8. #8
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    Quote Originally Posted by Plato View Post
    The standard way of proving the is to use properties of the complex conjugate.
    If a is a real number then \overline{a}=a.
    The conjugate of a sum is the sum of conjugates.
    Therefore we have:

    \overline {az^2  + bz + c}  = \overline 0  = 0\, \Rightarrow \,a\overline z ^2  + b\overline z  + c = 0
    ...and it is true for every polynomial...
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  9. #9
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    Fair point. So if I add:

    When b^2 - 4ac < 0, \sqrt{b^2 - 4ac} = ni, n \in R

    And replace the square root with the imaginary part 'n', would that suffice?
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  10. #10
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    Hello, Glitch!

    I have a very primitive solution . . .


    Use the properties of the complex conjugate to show that
    if the complex number z is a root of a quadratic equation
    f(x) \:=\:ax^2 + bx + c = 0 with a, b, c real coefficients,
    then so is the conjugate of z.

    We are told that z \:=\:p + qi is a solution of the quadratic.

    . . Hence: . a(p+qi)^2 + b(p+qi) + c \:=\:0

    This simplifies to: . (ap^2 - aq^2 + bp + c) + q(2ap + b)i \;=\;0

    . . And we have: . \begin{Bmatrix}ap^2 - aq^2 + bp + c \;=\;0 \\ q(2ap + b) \;=\;0 \end{Bmatrix}


    The conjugate of z is: . \overline z \:=\:p - qi

    Consider f(\overline z) \;=\;a(p-qi)^2 + b(p-qi) + c

    . . . . . . . f(\overline z) \;=\;\underbrace{(ap^2 - aq^2 + bp + c)}_{\text{This is 0}} - \underbrace{q(2ap + b)}_{\text{This is 0}}i


    \text{Therefore: }\:f(\overline z) \,=\,0\:\text{ and }\:\overline z\text{ is also a solution.}

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