Complex Numbers And Quadratic Equations question 656
Question: If $ 1,\omega ,{{\omega }^{2}} $ are the three cube roots of unity, then what is $ \frac{(a{{\omega }^{6}}+b{{\omega }^{4}}+c{{\omega }^{2}})}{(b+c{{\omega }^{10}}+a{{\omega }^{8}})} $ equal to?
Options:
A) $ \frac{a}{b} $
B) b
C) $ \omega $
D) $ {{\omega }^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
1, $ \omega $ and $ {{\omega }^{2}} $ are the three cube roots of unity.
$ \Rightarrow 1+\omega +{{\omega }^{2}}=0and{{\omega }^{3}}=1 $ The given expression $ \frac{a{{\omega }^{6}}+b{{\omega }^{4}}+c{{\omega }^{2}}}{b+c{{\omega }^{10}}+a{{\omega }^{8}}}=\frac{a+b\omega +c{{\omega }^{2}}}{b+c\omega +a{{\omega }^{2}}} $ $ [{{\omega }^{6}}=1,{{\omega }^{4}}=\omega ] $ $ =\frac{\omega (a+b\omega +c{{\omega }^{2}})}{\omega (b+c\omega +a{{\omega }^{2}})} $ $ [MultiplyingN^{r}andD^{r}by\omega ] $ $ =\frac{\omega (a+b\omega +c{{\omega }^{2}})}{(a{{\omega }^{3}}+b\omega +c{{\omega }^{2}})}=\frac{\omega (a+b\omega +c{{\omega }^{2}})}{(a+b\omega +c{{\omega }^{2}})}=\omega $
BETA
