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Math Help - 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 \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}
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    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.
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  3. #3
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    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.
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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 \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.
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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 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
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    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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