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Complex number

  1. Apr 17, 2004 #1
    u noe how x^5=1 has 5 roots which some of them are not real in complex field.

    and so is x^2=-64 with roots = -8i or 8i

    and i notice that the sum of roots = 0 (msut inculde non real --> complex number)

    is this becasue of the rule of polynomial --> -b/a = sum of roots

    for this case b always =0 so -b/a = 0 ?

    or (there is nothing to do with this and my example are just a fulke) once complex number is incolved then polynoimial rules can not apply?
  2. jcsd
  3. Apr 17, 2004 #2
  4. Apr 17, 2004 #3
    so this is current
  5. Apr 17, 2004 #4

    matt grime

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    what do you mean current? this result is well known, and I would suggest has been for very long time. The sum of all roots of x^n=1, n>1 (and be extension other numbers) is zero since, for example, they form the vertices of a regular n-gon.
  6. Apr 17, 2004 #5
    great i mean is this because of [tex]\frac{-b}{a}[/tex]

    sum of roots of a equation ax^n+bx^(n-1)+cx^(n-2)......

    this case x^2+0x^1 + 64=0 , x^2=-64

    sum of roots 0/a = 0
    Last edited: Apr 17, 2004
  7. Apr 17, 2004 #6

    matt grime

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    What's b what's a? The 'rules' about polynomials apply irrespective of the field, whereby I think you mean that for a monic polynomial of degree n the sum of the roots is the negative of the coeff of x^{n-1}

    your explanation appears retrospectively...
  8. Apr 17, 2004 #7
    ~~_~~~ sorry i m bad at explaning

    for example an equation of


    which a_{n} = a a_{n-1} = b

    so simmilar it can be write as

    this case

    x^2= -64

    a= 1 b= 0

    anyway wat i mean is that can rules of polynomial be applied to complex numbers

    omg i always confusing ppl how can i improve my explaning? help!!!
    Last edited: Apr 17, 2004
  9. Apr 17, 2004 #8

    matt grime

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    Why wouldn't the 'rules' of polynomials apply when the field is C?
  10. Apr 17, 2004 #9
    thx thx all cause my teacher said no and i doubt
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