# Complex number

1. Apr 17, 2004

### 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?

2. Apr 17, 2004

3. Apr 17, 2004

### expscv

so this is current

4. Apr 17, 2004

### 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.

5. Apr 17, 2004

### 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

Last edited: Apr 17, 2004
6. Apr 17, 2004

### 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}

7. Apr 17, 2004

### 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!!!

Last edited: Apr 17, 2004
8. Apr 17, 2004

### matt grime

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

9. Apr 17, 2004

### expscv

thx thx all cause my teacher said no and i doubt