# Thread: Sum of roots of a number

1. ## Sum of roots of a number

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 $\displaystyle \lambda$, such that $\displaystyle \lambda _k=\left | \lambda \right |^{1/n}exp^{2\pi ik/n} ...... k=0,1,2....n$ are the nth roots of $\displaystyle \lambda$, what is the sum of the roots,

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

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

2. Yes, it's $\displaystyle 0$, unless $\displaystyle n=1$!

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

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

4. Originally Posted by GeoC
So, are you saying the sum of the roots of $\displaystyle \lambda^{1/n}$ is 0 for n not equal 1, and, for n=1, the sum must be simply $\displaystyle \lambda$?
Yes!

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

5. ## Comment for Bruno J

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

7. Originally Posted by wonderboy1953