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
656.1k+ views
Hint: 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.
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.
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