Results 1 to 7 of 7

Math Help - complex numbers.pt2

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    25

    complex numbers.pt2

    Problem:

    Is it possible to determine the real constant a so that
    z=2+i
    is a root for the equation
    z^3+az^2+13z-10=0

    I started out like so that if p(b)=0, where p(z) is a polynomial of a degree n,
    then p(z)=(z-b)*f(z) where f(z) is a polynomial of a degree n-1

    meaning that

    z^3+az^3+13z-10=(z-(2+1))(z^2+dz+e)

    but apparently that was wrong..

    Sorry for all the questions, but I have an exam coming up this monday
    Last edited by Greenb; January 30th 2009 at 01:37 PM. Reason: Misstyping in equation
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Greenb View Post
    Problem:

    Is it possible to determine the real constant a so that
    z=2+1
    is a root for the equation
    z^3+az^3+13z-10=0

    I started out like so that if p(b)=0, where p(z) is a polynomial of a degree n,
    then p(z)=(z-b)*f(z) where f(z) is a polynomial of a degree n-1

    meaning that

    z^3+az^3+13z-10=(z-(2+1))(z^2+dz+e)

    but apparently that was wrong..

    Sorry for all the questions, but I have an exam coming up this monday
    Do you mean z = 2 + {\color{red}i} is the root?

    Just substitute into the cubic, simplify and equate the real and imaginary parts of the result to zero. Then solve for a.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    25
    This is what I got:

    (z-(2+i))(z^2+dz+e)=
    =z^3+dz^2+ez-2z^2-2dz-2e-iz^2-idz-ie

    somehow I think I need the values for d and e to proceed..

    How do I "equate the real and imaginary parts of the result to zero" seeing that I should leave the left side intact?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,788
    Thanks
    1683
    Awards
    1
    Please check your OP.
    It must contain some mistakes.
    You want 2+\color{red}i to be a root of z^3+az^{{\color{red}2}}+13z-10.
    If I have guessed correctly, there is a simple answer for a.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    25
    Ah didn't see that I had typed it wrong Anyway, yes a=-6 but I really can't see how.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,788
    Thanks
    1683
    Awards
    1
    Quote Originally Posted by Greenb View Post
    a=-6 but I really can't see how.
    Indeed, that is what I got.
    \left( {2 + i} \right)^3  + a\left( {2 + i} \right)^2  + 13\left( {2 + i} \right) - 10 = \left( {3a + 18} \right) + i\left( {4a + 24} \right)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2009
    Posts
    25
    I thought way more complicated and was actually totally off the track hehe. Thanks for clearing it up for me, I now see that mr fantastic also meant this solution, I was trying to fit in his advice into my solution ^^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: March 25th 2011, 10:02 PM
  2. Replies: 1
    Last Post: September 27th 2010, 03:14 PM
  3. Replies: 2
    Last Post: February 7th 2009, 06:12 PM
  4. Replies: 1
    Last Post: May 24th 2007, 03:49 AM
  5. Complex Numbers- Imaginary numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 24th 2007, 12:34 AM

Search Tags


/mathhelpforum @mathhelpforum