Complex Numbers And Quadratic Equations question 158
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 [MNR 1992; IIT 1984; DCE 2001; MP PET 2004]
Options:
A) 0
B) 1
C) $ n $
D) $ n^{2} $
Show Answer
Answer:
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 $
BETA
