Question

If $ {{a}_{1}},\,\,{{a}_{2}},...,\,\,{{a}_{n-1}} $ are the nth roots of unity, then the value of $ (1-{{a}_{1}})(1-{{a}_{2}})...(1-{{a}_{n-1}}) $ is equal to

• A. $ \sqrt{3} $
• B. $ \frac{1}{2} $
• C. $ n $
• D. $ 0 $

Answer 1

Answer: (C)

Since, $ 1,\,\,{{a}_{1}},\,\,{{a}_{2}},\,\,....,\,\,{{a}_{n-1}} $ are the nth root of unity.
$ \therefore $ $ {{x}^{n}}-1=(x-1)(x-{{a}_{1}})...(x-{{a}_{n-1}}) $
$ \Rightarrow $ $ \frac{{{x}^{n}}-1}{x-1}=(x-{{a}_{1}})(x-{{a}_{2}})...(x-{{a}_{n-1}}) $
$ \therefore $ $ {{x}^{n-1}}+{{x}^{n-2}}+...+{{x}^{2}}+x+1 $
$ =(x-{{a}_{1}})(x-{{a}_{2}})...(x-{{a}_{n-1}}) $
Put $ x=1 $ , we get $ (1-{{a}_{1}})(1-{{a}_{2}})...(1-{{a}_{n-1}})=1+1+...+n $ times $ =n $