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Sum of Roots of n-degree Polynomial


Date: 01/24/2001 at 22:36:54
From: Gary Schulte
Subject: Proving the n roots of unity sum to 0

I can demonstrate this to myself, but I can't find a way to prove it:

for all n roots of unity (z^n = 1) the sum of n roots is zero for 
n > 1.

Thanks for your help.
Gary


Date: 01/24/2001 at 22:46:34
From: Doctor Fenton
Subject: Re: Proving the n roots of unity sum to 0

Hi, Gary,

Thanks for writing to Dr. Math.  

Remember that the sum of the roots r(i) of a polynomial of degree n,
                                                    _n_
     a(n)*z^n + a(n-1)*z^(n-1)+...+ a(1)*z + a(0) = | | (z - r(i))
                                                    i=1
is a(n-1).

This result just comes from multiplying out the product. Any 
polynomial can be written as the product of the factors (z-ri), where 
the ri are its roots (and because of the fundamental Theorem of 
Algebra, every polynomial can be factored into linear factors over
the complex numbers). I'll divide by the leading coefficient a(n) so
that the first term is z^n.  For a quadratic,

      (z-r1)*(z-r2)=z^2 -(r1+r2)*z + r1*r2

and for a cubic,

  (z-r1)*(z-r2)*(z-r3) = z^3 -(r1+r2+r3)*z^2 +(r1*r2+r1*r3+r2*r3)*z 
                            -r1*r2*r3

The coefficients of such a product are called the "elementary 
symmetric polynomials" of the roots, and the coefficient of z^(n-1) is 
always -(r1+r2+...+rn). One way to see this is to notice that when 
computing the product of binomials

    (z-r1)*(z-r2)*...*(z-rn)

the product consists of the sum of all terms that can be formed by
taking one factor from each term in the product. If you take the z
from each term, you get z^n; if you take the constant from each 
factor, you get (-1)^n*(r1*r2*...*rn). To get a term with exponent 
z^(n-1), you must pick a z from each factor except one, and you have n 
choices of which factor to take the constant term from, so you will 
get the products

 -r1*z^(n-1)  (takes the constant from (z-r1), z from all other terms)

 -r2*z^(n-1)  (takes the constant from (z-r2), z from all other terms)

 -r3*z^(n-1)  (takes the constant from (z-r3), z from all other terms)
          :
 -rn*z^(n-1)  (takes the constant from (z-rn), z from all other terms)
   
Adding these up gives

 -(r1+r2+...+rn)*z^(n-1) .

So, for polynomial

     x^2 - 6x + 8

without solving I can tell that the sum of the roots is 6, and for 

    x^3 +3x^2 - 14x + 2

I know that the sum of the roots is -3.

Your roots of unity are roots of z^n - 1 = 0, so their sum must be 0.

You can find more information on the elementary symmetric polynomials
in most algebra books (e.g. Serge Lang's _Algebra_), or in Charles
Hadlock's book _Field Theory and Its Classical Problems_ or Ian 
Stewart's _Galois Theory_.  See also this answer in our Dr. Math 
archives:

  Symmetric Polynomials
  http://mathforum.org/dr.math/problems/stewart7.31.97.html   

If you have further questions, please write again.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Polynomials

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