Q55 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If $n$ is a positive integer, then $(1 + i\sqrt{3})^{2n} + (1 - i\sqrt{3})^{2n}$ is equal to .........
- Correct answerA$2^{2n+1}\cos\left(\dfrac{2n\pi}{3}\right)$
- B$2^{2n}\cos\left(\dfrac{n\pi}{3}\right)$
- C$2^{2n+1}\cos\left(\dfrac{n\pi}{3}\right)$
- D$2^n\cos\left(\dfrac{2n\pi}{3}\right)$
Solution
Let $z = 1 + i\sqrt{3}$.
Converting to polar form, we have $r = \sqrt{1^2 + (\sqrt{3})^2} = 2$ and $\theta = \tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}$.
Thus, $1 + i\sqrt{3} = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)$.
Similarly, $1 - i\sqrt{3} = 2\left(\cos\dfrac{\pi}{3} - i\sin\dfrac{\pi}{3}\right)$.
Using De Moivre's theorem, we get:
$(1 + i\sqrt{3})^{2n} = 2^{2n}\left(\cos\dfrac{2n\pi}{3} + i\sin\dfrac{2n\pi}{3}\right)$
$(1 - i\sqrt{3})^{2n} = 2^{2n}\left(\cos\dfrac{2n\pi}{3} - i\sin\dfrac{2n\pi}{3}\right)$
Adding the two expressions:
$(1 + i\sqrt{3})^{2n} + (1 - i\sqrt{3})^{2n} = 2^{2n}\left(2\cos\dfrac{2n\pi}{3}\right) = 2^{2n+1}\cos\dfrac{2n\pi}{3}$.
Answer: $2^{2n+1}\cos\left(\dfrac{2n\pi}{3}\right)$
Converting to polar form, we have $r = \sqrt{1^2 + (\sqrt{3})^2} = 2$ and $\theta = \tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}$.
Thus, $1 + i\sqrt{3} = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)$.
Similarly, $1 - i\sqrt{3} = 2\left(\cos\dfrac{\pi}{3} - i\sin\dfrac{\pi}{3}\right)$.
Using De Moivre's theorem, we get:
$(1 + i\sqrt{3})^{2n} = 2^{2n}\left(\cos\dfrac{2n\pi}{3} + i\sin\dfrac{2n\pi}{3}\right)$
$(1 - i\sqrt{3})^{2n} = 2^{2n}\left(\cos\dfrac{2n\pi}{3} - i\sin\dfrac{2n\pi}{3}\right)$
Adding the two expressions:
$(1 + i\sqrt{3})^{2n} + (1 - i\sqrt{3})^{2n} = 2^{2n}\left(2\cos\dfrac{2n\pi}{3}\right) = 2^{2n+1}\cos\dfrac{2n\pi}{3}$.
Answer: $2^{2n+1}\cos\left(\dfrac{2n\pi}{3}\right)$
Hence, the correct answer is option: 'A'.