Complex Numbers TG EAPCET Previous Year Chapters wise Questions

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Given the equation:$\;\left(\frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2}\right)^k + \left(\frac{a+b\omega+c\omega^2}{b+a\omega^2+c\omega}\right)^l = 2$where $\omega$ is the complex cube root of unity, meaning $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.To solve this, we first manipulate the terms by using properties of complex numbers. Consider multiplying and dividing:Multiply and divide the first term by $\omega^2$ :$\;\left(\frac{\omega^2 (a+b\omega+c\omega^2)}{\omega^2 (c+a\omega+b\omega^2)}\right)^k$Multiply and divide the second term by $\omega$ :$\;\left(\frac{\omega (a+b\omega+c\omega^2)}{\omega (b+a\omega^2+c\omega)}\right)^l$This simplifies to:$(\omega^{2 k}) + (\omega^l) = 2$For these powers to sum to 2 , and knowing the properties of cube roots of unity, we have specific implications for the exponents:Since $\omega^n$ can only take values $1, \omega$, or $\omega^2$, for the sum $\omega^{2 k} + \omega^l = 2$ to be a real number (i.e., $2)$, both $\omega^{2 k}$ and $\omega^l$ must equal 1 .Thus, this requires both 2 k and I to be multiples of $3 (.$ since $\omega^3 = 1)$. Hence, the expression $2 k+l$ is also a multiple of 3 , indicating that $2 k+l$ is always divisible by 3.