Complex Numbers And Quadratic Equations question 270
Question: If $ \alpha $ is an imaginary cube root of unity, then for $ n\in N $ , the value of $ {{\alpha }^{3n+1}}+{{\alpha }^{3n+3}}+{{\alpha }^{3n+5}} $ is [MP PET 1996; Pb. CET 2000]
Options:
A) $ -1 $
B) 0
C) 1
D) 3
Show Answer
Answer:
Correct Answer: B
Solution:
Since $ \alpha $ is an imaginary cube root of unity, let it be $ \omega $ then $ ={{(\omega )}^{3n+1}}+{{(\omega )}^{3n+3}}+{{\omega }^{3n+5}} $ $ =\omega +1+{{\omega }^{5}},{\because {{\omega }^{3n}}=1 $ and $ {{\omega }^{3}}=1} $ $ =\omega +1+{{\omega }^{2}}=0 $
BETA
