Complex Numbers And Quadratic Equations question 237
Question: If $ n $ is a positive integer not a multiple of 3, then $ 1+{{\omega }^{n}}+{{\omega }^{2n}} $ = [MP PET 2004]
Options:
A) 3
B) 1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ n=3k+1 $ $ {{\omega }^{n}}+{{\omega }^{2n}}={{\omega }^{3k+1}}+{{\omega }^{2(3k+1)}}={{\omega }^{3k}}\omega +{{\omega }^{6k}}{{\omega }^{2}} $ $ ={{({{\omega }^{3}})}^{k}}.\omega +{{({{\omega }^{3}})}^{2k}} $ . $ {{\omega }^{2}}=\omega +{{\omega }^{2}}=-1 $ Hence $ 1+{{\omega }^{n}}+{{\omega }^{2n}}=1-1=0 $
BETA
