Complex Numbers And Quadratic Equations question 164
Question: $ \omega $ is an imaginary cube root of unity. If $ {{(1+{{\omega }^{2}})}^{m}}= $ $ {{(1+{{\omega }^{4}})}^{m}}, $ then least positive integral value of m is [IIT Screening 2004]
Options:
6
5
4
3
Show Answer
Answer:
Correct Answer: D
Solution:
We have, $ {{(1+{{\omega }^{2}})}^{m}}={{(1+{{\omega }^{4}})}^{m}} $ $ (\because {{\omega }^{3}}=1) $ $ {{(1+{{\omega }^{2}})}^{m}}={{(1+{{\omega }^{2}})}^{m}} $ $ {{(-\omega )}^{m}}={{(-{{\omega }^{2}})}^{^{m}}} $
$ \Rightarrow {{( \frac{\omega }{{{\omega }^{2}}} )}^{m}}=1 $
$ \Rightarrow {{({{\omega }^{2}})}^{m}}=1 $ $ ={{(\omega )}^{2m}}=({{\omega }^{3m}}) $
$ \Rightarrow m=\frac{3}{2} $ Hence least positive integral value of m is 2.
BETA
