Question

If w is a complex cube root of unity, then the value of

$\frac{a + b\omega + c\omega^{2}}{c + a\omega + b\omega^{2}} + \frac{a + b\omega + c\omega^{2}}{b + c\omega + a\omega^{2}}$ is

• A. 1
• B. 0
• C. 2
• D. – 1

Answer 1

Answer: (D)

– 1

Answer 2

Sol. $\frac{a + b\omega + c\omega^{2}}{a\omega + b\omega^{2} + c} + \frac{a + b\omega + c\omega^{2}}{b + c\omega + a\omega^{2}}$

= $\frac{1}{\omega}\left( \frac{a\omega + b\omega^{2} + c\omega^{3}}{a\omega + b\omega^{2} + c} \right) + \frac{1}{\omega^{2}}\left( \frac{a\omega^{2} + b\omega^{3} + c\omega^{4}}{a\omega^{2} + b + c\omega} \right)$

= w + w² = 1 (Q 1/w = w² , Q w³ = 1)