Results 1 to 6 of 6

Math Help - Complex Numbers, Find The Unknown - Help

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    16

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

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

    Let z=-2+2i=2(-1+i).

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

    We know that P(z)=0.

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

    Because you know all the coefficients of 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, fz^2+gz+8=0.

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

    2ig-2g-8if+8=0

    (-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 :

    \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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
    16
    Thanks, Moo! That helped me a lot

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

    - Matt.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    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 (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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by Phatmat View Post
    I am stuck on question 11, it is below:
    We have

    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 P(x) = 0. Also note that the nonreal roots of a polynomial with real coefficients always come in conjugate pairs. So if -2 + 2{\rm i} is a root, then so must -2 - 2{\rm i}.

    We may easily find the other two roots thus:

    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 \left(fx^2 + gx + 8\right) factor. So we have

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

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

    Equating coefficients, we find that C = 1, so

    f = 1, g = 4

    Then we have

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

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

    \Rightarrow\left\{\begin{array}{rcl}<br />
a & = & 5\\<br />
b & = & 26\\<br />
c & = & 69\\<br />
d & = & 68\\<br />
e & = & 40<br />
\end{array}\right.

    If you like, you can do the substitutions and verify that these are the correct values.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2008
    Posts
    16
    Thanks for the help, those are the correct answers
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Numbers - Find z & w
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 17th 2011, 07:10 AM
  2. Find all complex numbers satisfying z^6=1
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: March 17th 2010, 08:31 PM
  3. find complex numbers
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 7th 2010, 08:56 PM
  4. Complex numbers/find a and b
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 21st 2009, 04:59 AM
  5. Complex Numbers...Find x
    Posted in the Algebra Forum
    Replies: 7
    Last Post: December 29th 2008, 11:02 PM

Search Tags


/mathhelpforum @mathhelpforum