Complex number

  • Thread starter expscv
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  • #1
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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?
 

Answers and Replies

  • #3
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so this is current
 
  • #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.
 
  • #5
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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:
  • #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...
 
  • #7
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~~_~~~ sorry i m bad at explaning

for example an equation of

[tex]a_{n}x^{n}+a_{n-1}x^{n-1}+......+a_{1}x+a_{0}[/tex]

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

so simmilar it can be write as
[tex]ax^n+bx^{n-1}+....[/tex]

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:
  • #8
matt grime
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Why wouldn't the 'rules' of polynomials apply when the field is C?
 
  • #9
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thx thx all cause my teacher said no and i doubt
 

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