Are the Sum of Polynomial Roots Always Zero When Complex Numbers Are Involved?

[expscv]

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?

 

[cookiemonster]

Vieta's Formula is the general case. Mathworld has a pretty fair treatment of it.

http://mathworld.wolfram.com/PolynomialRoots.html
http://mathworld.wolfram.com/VietasFormulas.html

cookiemonster

 

[expscv]

so this is current

 

[matt grime]

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.

 

[expscv]

great i mean is this because of \frac{-b}{a}

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

 

[matt grime]

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...

 

[expscv]

~~_~~~ sorry i m bad at explaning

for example an equation of

a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}

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

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

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!

 

[matt grime]

Why wouldn't the 'rules' of polynomials apply when the field is C?

 

[expscv]

thx thanks all cause my teacher said no and i doubt