Complex Numbers And Quadratic Equations question 262
Question: $ (1-\omega +{{\omega }^{2}})(1-{{\omega }^{2}}+{{\omega }^{4}})(1-{{\omega }^{4}}+{{\omega }^{8}})……….. $ to $ 2n $ factors is [EAMCET 1988]
Options:
A) $ 2^{n} $
B) $ 2^{2n} $
C) 0
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\omega }^{4}}=\omega ,{{\omega }^{8}}={{\omega }^{2}} $ etc. 3rd, 5th, 7th factors are each equal to 1st and 4th, 6th, 8th factors are each equal to 2nd. L.H.S $ =(-2\omega )(-2{{\omega }^{2}})(-2\omega )(-2{{\omega }^{2}})….. $ to $ 2n $ factors $ =(2^{2}{{\omega }^{3}})(2^{2}{{\omega }^{3}})…… $ to $ n $ factors $ ={{(2^{2})}^{n}}=2^{2n} $
BETA
