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Let $\omega$ denote a complex fifth root of unity. Define $$b_k =\sum_{j=0}^{4} j \omega^{-kj},$$ for $0 \leq k \leq 4$. Then $ \sum_{k=0}^{4} b_k \omega ^k$ is equal to

  1. $5$
  2. $5 \omega$
  3. $5(1+\omega)$
  4. $0$
in Others by Veteran (425k points)
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1 Answer

+1 vote

Here, $\omega^5=1\Rightarrow \omega^4=\omega^{-1}\Rightarrow \omega^3=\omega^{-2}\Rightarrow \omega^2=\omega^{-3}$

$\therefore \omega^{5-k}=\omega^{-k}$ and $\omega^{5n}=1,~ n \in \mathbb{Z}$.

Now

$\begin{align} \sum_{k=0}^{4}b_k \omega^{k} &= \sum_{k=0}^{4}\left( \sum_{j=0}^{4}j \omega^{-kj} \right) \omega^{k} \\ &=\sum_{k=0}^{4}\sum_{j=0}^{4}j \omega^{k-kj}\\ &=\sum_{k=0}^{4} \left(0+ \omega^{0}+2 \omega^{-k}+ 3\omega^{-2k}+ 4\omega^{-3k} \right)\\ &=\sum_{k=0}^{4} \left(1+2 \omega^{4k}+ 3\omega^{3k}+ 4\omega^{2k} \right)\\ &=\sum_{k=0}^{4}1+2\sum_{k=0}^{4}(\omega^{4})^{k}+3\sum_{k=0}^{4}(\omega^{3})^{k}+4\sum_{k=0}^{4}(\omega^{2})^{k}\\ &=5+2\frac{(\omega^4)^5-1}{\omega^4-1}+3\frac{(\omega^3)^5-1}{\omega^3-1}+4\frac{(\omega^2)^5-1}{\omega^2-1}\\ &=5+2\frac{\omega^{20}-1}{\omega^4-1}+3\frac{\omega^{15}-1}{\omega^3-1}+4\frac{\omega^{10}-1}{\omega^2-1}\\ &=5+0+0+0; ~[\because \omega^{5n}=1]\\ &=5 \end{align}$

 

So the correct answer is A.

by Active (3.2k points)
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Classic one :D

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