A) 0
B) 1
C) \[n\]
D) \[{{n}^{2}}\]
Correct Answer: C
Solution :
Since\[1,\omega ,{{\omega }^{2}},{{\omega }^{3}},.....{{\omega }^{n-1}}\]are the \[n,{{n}^{th}}\] roots of unity, therefore, we have the identity \[=(x-1)(x-\omega )(x-{{\omega }^{2}}).....(x-{{\omega }^{n-1}})={{x}^{n}}-1\] or \[(x-\omega )(x-{{\omega }^{2}}).....(x-{{\omega }^{n-1}})=\frac{{{x}^{n}}-1}{x-1}\] =\[{{x}^{n-1}}+{{x}^{n-2}}+.....+x+1\] Putting \[x=1\] on both sides, we get \[(1-\omega )(1-{{\omega }^{2}}).....(1-{{\omega }^{n-1}})=n\]
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