Sum of roots of a number

May 2010
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1
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. \(\displaystyle \left | \lambda \right |^{1/n}\sum_{k=0}^{n-1}exp^{2\pi ik/n}\)
 

Bruno J.

MHF Hall of Honor
Jun 2009
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Canada
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 \(\displaystyle f(z)=z^n-\lambda\).
 
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May 2010
43
1
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.
 

Bruno J.

MHF Hall of Honor
Jun 2009
1,266
498
Canada
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 \(\displaystyle S=0\).
 
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chisigma

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Mar 2009
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near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)