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Thread: Complex Numbers, Find The Unknown - Help

  1. #1
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    Complex Numbers, Find The Unknown - Help

    I am stuck on question 11, it is below:



    If anyone is able to do this it would be great, but if you can somehow show me the working out/steps involved. Though, any tips/help would also be appreciated.

    thanks,
    Matt.

    EDIT: Sorry! I posted in the wrong section!
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  2. #2
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    Hello,

    $\displaystyle \begin{aligned}P(x)&=ax^4+bx^3+cx^2+dx+e \\
    &=(5x^2+6x+5)(fx^2+gx+8) \end{aligned}$

    Let $\displaystyle z=-2+2i=2(-1+i)$.

    ----------------

    We know that $\displaystyle P(z)=0$.

    So $\displaystyle (5z^2+6z+5)(fz^2+gz+8)=0$.

    Because you know all the coefficients of $\displaystyle 5z^2+6x+5$, calculate if it is equal to 0.
    After some calculations (that you will do), it seems that it is not equal to 0.

    Therefore, $\displaystyle fz^2+gz+8=0$.

    Substituting z by its value, and after some calculations (that you will do), we get :

    $\displaystyle 2ig-2g-8if+8=0$

    $\displaystyle (-2g+8)+i(2g-8f)=0$

    This is a complex number. But a complex number equals to 0 if and only if the real and the imaginary parts are both equal to 0, that is to say :

    $\displaystyle \left\{ \begin{array}{ccc}-2g+8&=&0 \\ 2g-8f&=&0 \end{array}\right.$

    This gives f and g

    --------------

    I have to go out right now, so if no one else does, I'll help you find a,b,c,d, and e
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    Thanks, Moo! That helped me a lot

    I have solved a,b,c,d,e my self, thanks again for the help!

    - Matt.
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    Quote Originally Posted by Phatmat View Post
    I am stuck on question 11, it is below:



    If anyone is able to do this it would be great, but if you can somehow show me the working out/steps involved. Though, any tips/help would also be appreciated.

    thanks,
    Matt.

    EDIT: Sorry! I posted in the wrong section!
    Since the coefficients of p(x) are real, z = 2 - 2i is also a root by the conjugate root theorem.

    Hence a quadratic factor of p(x) is $\displaystyle (x - [-2 + 2i])(x - [-2 - 2i]) ([x + 2] - 2i)([x + 2] + 2i) = (x+2)^2 + 4 = x^2 + 4x + 8$.

    So you now have f = 1 and g = 4 in the second expression. Now expand the second expression to get the coefficients a, b, c, d, e in the first.
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    Quote Originally Posted by Phatmat View Post
    I am stuck on question 11, it is below:
    We have

    $\displaystyle P(x) = ax^4 + bx^3 + cx^2 + dx + e = \left(5x^2 + 6x + 5\right)\left(fx^2 + gx + 8\right),\;a,b,c,d,e,f,g\in\mathbb{R}$

    First, note that a polynomial of degree 4 has at most four distinct roots (this follows from the fundamental theorem of algebra), so we should get at most four distinct solutions to the equation $\displaystyle P(x) = 0$. Also note that the nonreal roots of a polynomial with real coefficients always come in conjugate pairs. So if $\displaystyle -2 + 2{\rm i}$ is a root, then so must $\displaystyle -2 - 2{\rm i}$.

    We may easily find the other two roots thus:

    $\displaystyle 5x^2 + 6x + 5 = 0\Rightarrow x = \frac{-3\pm4{\rm i}}5$

    This means that the first two roots we identified must come from the $\displaystyle \left(fx^2 + gx + 8\right)$ factor. So we have

    $\displaystyle \left(fx^2 + gx + 8\right) = C\left(x + 2 - 2{\rm i}\right)\left(x + 2 + 2{\rm i}\right)$

    $\displaystyle = C\left(x^2 + 4x + 8\right)$ for some $\displaystyle C\in\mathbb{R}$

    Equating coefficients, we find that $\displaystyle C = 1$, so

    $\displaystyle f = 1, g = 4$

    Then we have

    $\displaystyle ax^4 + bx^3 + cx^2 + dx + e = \left(5x^2 + 6x + 5\right)\left(x^2 + 4x + 8\right)$

    $\displaystyle =5x^4+26x^3+69x^2+68x+40$

    $\displaystyle \Rightarrow\left\{\begin{array}{rcl}
    a & = & 5\\
    b & = & 26\\
    c & = & 69\\
    d & = & 68\\
    e & = & 40
    \end{array}\right.$

    If you like, you can do the substitutions and verify that these are the correct values.
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  6. #6
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    Thanks for the help, those are the correct answers
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