• # question_answer If $1,\omega ,{{\omega }^{2}},{{\omega }^{3}}.......,{{\omega }^{n-1}}$ are the $n,{{n}^{th}}$ roots of unity, then $(1-\omega )(1-{{\omega }^{2}}).....(1-{{\omega }^{n-1}})$ equals [MNR 1992; IIT 1984; DCE 2001; MP PET 2004] A) 0 B) 1 C) $n$ D) ${{n}^{2}}$

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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