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Problem with polynomial division and 'i'

  1. Jun 20, 2004 #1
    I was learning polynomial division, and I can do most problems, except this one which is bothering me.

    :

    3x^2 + 2x + 7
    ---------------
    (1+i)x - 2


    How would I divide something like that? Nothing is working. Thanks.
     
  2. jcsd
  3. Jun 20, 2004 #2
    [tex]i[/tex] is just another number, no different from 12 or [tex]pi[/tex] or [tex]\sqrt{2}[/tex]... You'd perform the long division like you usually do. The first term in the quotient is [tex]\frac{3}{1 + i}x[/tex], maybe that can get you started.
     
  4. Jun 20, 2004 #3

    HallsofIvy

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    Another way to handle fractions involving compex numbers is to "realize" the denominator. (I just made up that word!) Multiply both numerator and denominator by the complex conjugate of the denominator: the complex conjugate of (1+i)x - 2 is (1-i)x- 2 (negative i instead of positive i). Multiplying the denominator (and numerator) by that gives you a fraction in which the numerator is a real number.
     
  5. Jun 20, 2004 #4

    jcsd

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    You really want to mutiply through by the complex conjugate of 1+i tho'


    edited to add Halls of Ivy beat me to it, again!
     
  6. Jun 20, 2004 #5
    I tried the problem and I got

    3x - 1 ...... 3xi-i+6
    ------ remainder --------
    1+ i ....... (1+i)x-2



    Anybody know where I'm going wrong?
     
  7. Jun 20, 2004 #6

    jcsd

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    What you want to do is mutply the top and bottom by he complex conjugate of 1 + i which is 1 - i to get:

    [tex]\frac{(1-i)(3x^2 + 2x + 7)}{2x -4}[/tex]
     
  8. Jun 20, 2004 #7
    I did that, and now I have a [tex]1-i[/tex] in the dividend, so how do I divide by that term using polynomial long division (should I multiply it by [tex]3x^2+2x+7[/tex] ?)
     
  9. Jun 20, 2004 #8
    I get [tex]\frac{3}{2}x + 4 + \frac{23}{(2x-4)}[/tex], but where does the [tex]1-i[/tex] come in?
     
  10. Jun 20, 2004 #9
    Is the denominator (1+i)(x-2) or (1+i)x-2 ?
     
  11. Jun 20, 2004 #10
    The denominator is [tex](1+i)x-2[/tex]
     
  12. Jun 21, 2004 #11

    HallsofIvy

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    [tex]\frac{3x^2+ 2x+ 7}{(1+i)x- 2}[/tex]

    The denominator is (1+i)x- 2 and its complex conjugate is (1-i)x- 2 (just replace i by -i).

    Multiply both numerator and denominator by (1-i)x- 2

    The numerator will become (1-i)x(3x2+ 2x+ 7)- 2(3x2+ 2x+ 7)
    = (3-3i)x3+ (2-2i)x2+ (7-7i)x- 6x2- 4x- 14
    = (3-3i)x3+ (-4-2i)x2+(3- 7i)x- 14

    The denominator will become (1-i)x(1+i)x-2(1+i)x-2(1-i)x+ 4
    = 2x2- 4x+ 4 (No i !!)

    so the fraction is [tex]\frac{(3-3i)x^3+ (-4-2i)x^2+ (3-7i)x- 14}{2x^2- 4x- 4} [/tex]

    Now use long division to reduce that.
     
  13. Jun 21, 2004 #12
    Of what use is polynomial long division (other than to find the leading-order behavior of a rational polynomial)?
     
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