Sum of roots of a number

**GeoC** · May 27th 2010, 12:18 PM

Is there a theorem related to the sum of the roots of a positive number?

For example, for any integer n=1,2,3.....N, and positive number $\lambda$ , such that $\lambda _k=\left | \lambda \right |^{1/n}exp^{2\pi ik/n} ...... k=0,1,2....n$ are the nth roots of $\lambda$ , what is the sum of the roots,

i.e. $\sum_{k=0}^{n-1}\lambda _k = ?$ .

In other words, is there a closed form, and general solution for the sum of roots, i.e. $\left | \lambda \right |^{1/n}\sum_{k=0}^{n-1}exp^{2\pi ik/n}$

**Bruno J.** · May 27th 2010, 12:35 PM

Yes, it's $0$ , unless $n=1$ !

Hint : the sum of the roots of a polynomial $f(z)$ is (up to sign) the coefficient of $z$ , and the roots of a complex number $\lambda$ are the roots of $f(z)=z^n-\lambda$ .

**GeoC** · May 27th 2010, 03:44 PM

So, are you saying the sum of the roots of $\lambda^{1/n}$ is 0 for n not equal 1, and, for n=1, the sum must be simply $\lambda$ ? Thanks.. looking at the $\sum$ , I guess that makes sense.. I had long since forgotten, if I ever even knew it, that the sum of roots equals coefficient of Z in the polynomial.

**Bruno J.** · May 27th 2010, 10:37 PM

Originally Posted by GeoC

So, are you saying the sum of the roots of $\lambda^{1/n}$ is 0 for n not equal 1, and, for n=1, the sum must be simply $\lambda$ ?

Yes!

Here's another way to prove it : if $n>1$ , let $\omega = e^{2\pi i / n}$ and $S=1+\omega + \dots + \omega^{n-1}$ . Then, since $\omega^n=1$ , we have $\omega S = \omega + \dots + \omega^n = S$ . Since $\omega \neq 1$ , we must have $S=0$ .

**wonderboy1953** · May 28th 2010, 11:58 AM

I like your new Avatar.

**chisigma** · May 28th 2010, 12:53 PM

In general if we have a polynomial of degree $n$ in $z$ it can be written as...

$p(z) = z^{n} + a_{n-1} z^{n-1} + \dots + a_{1} z + a_{0} = \prod_{k=0}^{n-1} (z-z_{k})$ (1)

... where the $z_{k}$ , $k=0,1,\dots , n-1$ are the roots of the polynomial. If the roots are all distinct then from (1) is easy to derive that is...

$\sum_{k=0}^{n-1} z_{k} = - a_{n-1}$ (2)

If $p(z) = z^{n} - \lambda$ with $\lambda>0$ is...

$\sum_{k=0}^{n-1} z_{k} = \sum_{k=0}^{n-1} \lambda^{\frac{1}{n}} e^{2 \pi i \frac{k}{n}} =0$ (3)

Kind regards

$\chi$ $\sigma$

**Bruno J.** · May 28th 2010, 02:17 PM

Originally Posted by wonderboy1953

I like your new Avatar.

Thanks! Do you have an idea what it's of?

**wonderboy1953** · June 1st 2010, 06:50 AM

Originally Posted by Bruno J.

Thanks! Do you have an idea what it's of?

Taking a stab.

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