Associated Topics || Dr. Math Home || Search Dr. Math

### 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.

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)

-(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.

in most algebra books (e.g. Serge Lang's _Algebra_), or in Charles
Hadlock's book _Field Theory and Its Classical Problems_ or Ian
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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search