Complex Numbers And Quadratic Equations question 295
Question: If $ \omega $ is a complex cube root of unity, then the value of $ {{\omega }^{99}}+{{\omega }^{100}}+{{\omega }^{101}} $ is [Pb. CET 2004]
Options:
A) 1
B) - 1
C) 3
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{\omega }^{99}}+{{\omega }^{1\omega }}+{{\omega }^{101}} $ = $ {{\omega }^{99}}[1+\omega +{{\omega }^{2}}] $ [Since $ 1+\omega +{{\omega }^{2}}=0,{{\omega }^{3}}=1 $ ] = 0.
BETA
