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Math Help - Sum over roots of unity

  1. #1
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    Sum over roots of unity

    Let \omega be a real or complex number satisfying \omega^q = 1, where q is a natural number. Show that \sum_{\omega} \omega^k has the value q if k is divisible by q and has the value 0 otherwise.

    Slightly annoying that I cannot prove this. The case k is divisible by q is obvious. I can only prove the sum is equal to 0 when (k,q)=1...
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  2. #2
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    Quote Originally Posted by Capillarian View Post
    Let \omega be a real or complex number satisfying \omega^q = 1, where q is a natural number. Show that \sum_{\omega} \omega^k has the value q if k is divisible by q and has the value 0 otherwise.

    Slightly annoying that I cannot prove this. The case k is divisible by q is obvious. I can only prove the sum is equal to 0 when (k,q)=1...
    .


    Divide with residue k=xq+r\,,\,\,r=0\,\,or\,\,|r|<q , so

    \displaystyle{\sum\limits_ww^k=\sum\limits_w(w^q)^  xw^r=\sum\limits_ww^r .

    As you said, if k is divisible by q then r = 0 and the sum is trivially = q, otherwise:

    Take a q-th root of unity \sigma s.t. \sigma^r\neq 1 (why must there be such a root?), so:

    \displaystyle{\sigma^r\sum\limits_ww^r=\sum\limits  _w\sigma^rw^r=\sum\limits_w(\sigma w)^r=\sum\limits_ww^r ,

    the last equality fowllowing because... ,and then

    \displaystyle{\sigma^r\sum\limits_ww^r=\sum\limits  _ww^r\Longrightarrow (\sigma^r-1)\sum\limits_ww^r=0 , and we're done.

    Tonio

    Pd. The above proof assumes you know a little group theory and that the q-th roots

    of unity are a (cyclic) a group under complex multiplication.
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