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Thread: Sum of roots of a number

  1. #1
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    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. $\displaystyle \left | \lambda \right |^{1/n}\sum_{k=0}^{n-1}exp^{2\pi ik/n}$
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    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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  3. #3
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    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.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by GeoC View Post
    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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  5. #5
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    Comment for Bruno J

    I like your new Avatar.
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    MHF Contributor chisigma's Avatar
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    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

    $\displaystyle \chi$ $\displaystyle \sigma$
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by wonderboy1953 View Post
    I like your new Avatar.
    Thanks! Do you have an idea what it's of?
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  8. #8
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    MS Escher

    Quote Originally Posted by Bruno J. View Post
    Thanks! Do you have an idea what it's of?
    Taking a stab.
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