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Question: If \(\omega\)is a complex cube root of unity, then value of expression \(\cos\left\lbrack \left\{ (1...

If ω\omegais a complex cube root of unity, then value of expression cos⁡[{(1−ω)(1−ω)2+......+(10−ω)(10−ω2)}π900]\cos\left\lbrack \left\{ (1 - \omega)(1 - \omega)^{2} + ...... + (10 - \omega)\left( 10 - \omega^{2} \right) \right\}\frac{\pi}{900} \right\rbrack

A

-1

B

0

C

1

D

32\frac{\sqrt{3}}{2}

Answer

0

Explanation

Solution

Sol. We have (k−ω)(k−ω)2=k2−k(ω+ω2)+ω3(k - \omega)(k - \omega)^{2} = k^{2} - k\left( \omega + \omega^{2} \right) + \omega^{3}

= k2−k(−1)+1=k2+k+1k^{2} - k( - 1) + 1 = k^{2} + k + 1

⇒ ∑k=110(k−ω)(k−ω)2=∑k=110(k2+k+1)\sum_{k = 1}^{10}{(k - \omega)(k - \omega)^{2}} = \sum_{k = 1}^{10}\left( k^{2} + k + 1 \right)

=10 x 11 x 216+10 x 112+10=385+55+10=450\frac{10\text{ x 11 x 21}}{6} + \frac{10\text{ x 11}}{2} + 10 = 385 + 55 + 10 = 450Thus, cos⁡[{∑k=110(k−ω)(k−ω)2}π900]=cos⁡(450900π)=0\cos\left\lbrack \left\{ \sum_{k = 1}^{10}{(k - \omega)(k - \omega)^{2}} \right\}\frac{\pi}{900} \right\rbrack = \cos\left( \frac{450}{900}\pi \right) = 0.