Question

If $\omega$ is complex cube root of unity than the value of $\dfrac{{a + b\omega + c{\omega ^2}}}{{c + a\omega + b{\omega ^2}}} + \dfrac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}}$
A. $0$
B. $- 1$
C. $1$
D. $2$

Answer 1

As we know that the complex of unity is $\omega$ , so some other values are $1 + \omega + {\omega ^2} = 0$ and ${\omega ^3} = 1$ . So using all these values and we first, multiply the first term with ${\text{omega }}$ to both numerator and denominator and then on further simplification we get our answer.

Complete step by step answer:

Considering the given equation, $\dfrac{{a + b\omega + c{\omega ^2}}}{{c + a\omega + b{\omega ^2}}} + \dfrac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}}$
Let first multiply and divide the first term with ${\text{omega }}$ , we get,

$$= \dfrac{{(a + b\omega + c{\omega ^2})\omega }}{{(c + a\omega + b{\omega ^2})\omega }} + \dfrac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}} \\\ = \dfrac{{a\omega + b{\omega ^2} + c{\omega ^3}}}{{c\omega + a{\omega ^2} + b{\omega ^3}}} + \dfrac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}} \\\$$

Now as ${\omega ^3} = 1$ , we get,
$= \dfrac{{a\omega + b{\omega ^2} + c}}{{(c\omega + a{\omega ^2} + b)}} + \dfrac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}}$
As denominator are same we can add them directly and so, we get,
$= \dfrac{{a(1 + \omega ) + b(\omega + {\omega ^2}) + c(1 + {\omega ^2})}}{{(b + c\omega + a{\omega ^2})}}$
Now, using $1 + \omega + {\omega ^2} = 0$ so simplify various terms of numerator, we get,
$= \dfrac{{a( - {\omega ^2}) + b( - 1) + c( - \omega )}}{{(b + c\omega + a{\omega ^2})}}$
Taking minus common we get,

$$= \dfrac{{ - (b + c\omega + a{\omega ^2})}}{{(b + c\omega + a{\omega ^2})}} \\\ = - 1 \\\$$

Hence, option (B) is our correct answer.

Note: The cube roots of unity can be defined as the numbers which when raised to the power of 3 will give the result as 1. A complex number is a number that can be expressed in the form $a{\text{ }} + {\text{ }}bi$ , where a and b are real numbers, and i represents the imaginary unit, satisfying the equation ${i^2} = - 1$. Because no real number satisfies this equation, i is called an imaginary number.