Question

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

• A. \N
• B. 1
• C. n
• D. $n^2$

Answer 1

The correct option is C n

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
$(1-\omega )(1-{\omega }^2)....(1-{\omega }^{n-1})=n$