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Complex numbers, solving polynomial, signs of i

  1. Jun 7, 2010 #1
    I'm revising complex numbers and having trouble with this question...

    Question:

    Verify that 2 of the roots of the equation:

    z^3 +2z^2 + z + 2 = 0

    are i and -2. Find any remaining roots

    Attempt at solution:

    i^3 +2 i^2 + i + 2 =
    (-1)i + 2(-1) +i + 2 =
    -i -2 + i +2 =0
    therefore i is a root

    (-2)^3 +2(-2)^2 + (-2) + 2 =
    -8 + 8 = 0
    therefore -2 is a root


    Let y be the remaining root to be found...

    (z+i) (z-2)(z+y) = 0
    (z^2 - 2z + iz -2i) (z+y) = 0
    z^3 - 2z^2 + iz^2 -2iz +z^2 y -2zy +izy -2iy = 0

    z^3 + (i+y-2) z^2 + (iy-2y-2i)z + (-2iy) = 0

    so from this I should be able to work out y by equating the coefficients...
    i + y-2 = 2 -> i+y = 4??
    iy-2-2i = 1 iy - 2i = 3?
    -2iy = 2 from this i think y=i but it doesn't seem to agree with the other equations.

    This is where i'm stuck.
    Does i multiplied by -i equal 1 or -1?
    Also what is is (-i)^2 ?

    Thanks
     
  2. jcsd
  3. Jun 7, 2010 #2

    CompuChip

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    Science Advisor
    Homework Helper

    I think the mistake is at the very beginning...
    (z+i) (z-2)(z+y) = 0

    The idea of factoring is that if you plug in a root, one of the factors will be zero. Therefore, you should always write (z - root). If +i and -2 are roots, then, you would get
    (z-i)(z+2)(z-y) = 0

    i * -i = -1 * i *i = -1 * -1 = 1

    (-i)^2 = i^2 = -1
    just like
    (-2)^2 = 2^2 = 4
     
  4. Jun 7, 2010 #3

    Mark44

    Staff: Mentor

    Following up on what Compuchip said, if i and -2 are roots, z - i and z + 2 are factors, which means also that (z - i)(z + 2) is a factor. If you multiply this out and divide your original polynomial by it, you will get your third factor.
     
  5. Jun 7, 2010 #4

    mathman

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    Science Advisor

    One thing that you should always be aware of is that if all the coefficients of a polynomial are real, then the complex roots will come in conjugate pairs. The conjugate of i is -i.
     
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