Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\sum_{k=0}^4b_k\omega^{k}$?

share|cite|improve this question
up vote 3 down vote accepted

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, $B=1\cdot s(1)=5$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.