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Let $\omega$ denote a complex fifth root of unity. Define

$b_k = \sum_{j=0}^4j\omega^{-kj}$

for $0\le k\le 4$.

What is the value of $\sum_{k=0}^4b_k\omega^{k}$?

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The answer is $B=\sum\limits_{j=0}^4j\cdot s(\omega^{j-1})$ with $s(z)=\sum\limits_{k=0}^4z^{-k}$. If $z=\omega^{j-1}$ for some $j$ then $z^5=1$. If $z^5=1$ and $z\ne1$, then $s(z)=0$. Hence $s(\omega^{j-1})=0$ except if $j=1$. Thus, $B=1\cdot s(1)=5$.

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Perfect. But is it $\omega^{j-1}$ or $\omega^{1-j}$? –  user2204800 Apr 24 '13 at 16:27
    
Now it is $\omega^{j-1}$. Funny: you do not upvote the answers you accept? –  Did Apr 28 '13 at 13:25

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