Q55 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
The value of $\left(\dfrac{-1+i\sqrt{3}}{2}\right)^{18} + \left(\dfrac{-1-i\sqrt{3}}{2}\right)^{18}$ is
- Correct answerA$2$
- B$35$
- C$\sqrt{3}$
- D$\dfrac{\sqrt{3}}{2}$
Solution
Let $\omega = \dfrac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \dfrac{-1-i\sqrt{3}}{2}$ be the complex cube roots of unity.
The given expression can be written as $\omega^{18} + (\omega^2)^{18}$.
Since $\omega^3 = 1$, we have $\omega^{18} = (\omega^3)^6 = 1^6 = 1$.
Similarly, $(\omega^2)^{18} = \omega^{36} = (\omega^3)^{12} = 1^{12} = 1$.
Therefore, the value of the expression is $1 + 1 = 2$.
Answer: $2$
The given expression can be written as $\omega^{18} + (\omega^2)^{18}$.
Since $\omega^3 = 1$, we have $\omega^{18} = (\omega^3)^6 = 1^6 = 1$.
Similarly, $(\omega^2)^{18} = \omega^{36} = (\omega^3)^{12} = 1^{12} = 1$.
Therefore, the value of the expression is $1 + 1 = 2$.
Answer: $2$
Hence, the correct answer is option: 'A'.