Complex Numbers And Quadratic Equations question 278
If $ e^{i\pi /3} $ is a complex root of the equation $ z^{3}=1 $ , then $ \omega +{{\omega }^{( \frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+… )}} $ is equal to [Roorkee 2000; AMU 2005]
Options:
- 1
0
9
i
Show Answer
Answer:
Correct Answer: A
Solution:
$ \omega +{{\omega }^{( \frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+….. )}} $ $ =\frac{(1-\cos \theta )-i\sin \theta }{{{(1-\cos \theta )}^{2}}+{{\sin }^{2}}\theta } $ $ \omega +{{\omega }^{( \frac{1/2}{1-3/4} )}} $
$ \Rightarrow $ $ \omega +{{\omega }^{2}}=-1 $ $ [\because 1+\omega +{{\omega }^{2}}=0] $
BETA
