If n is a positive integer, then (1+i {3})² n}+(1-i {3})² n} is equal to \_\_\_\_

Correct Answer is : 22n+1cos⁡(2nπ3)2² n+1} ≤ft(2 n π/3)22n+1cos(32nπ)

Correct Answer is : 22n+1cos⁡(2nπ3)2² n+1} ≤ft(2 n π/3)22n+1cos(32nπ)

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MHTCET 16 Apr 2026 Shift 2 Paper

Question : 101 of 150

Marks: +1, -0

n

(1+i \sqrt{3})^{2 n}+(1-i \sqrt{3})^{2 n}

\_\_\_\_

Solution:

Concept:

Use polar form of complex numbers and De Moivre's theorem to simplify the sum.

Explanation:

For

1 + i\sqrt{3}

, the modulus is

|1 + i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = 2

Its argument is given by

\tan\theta = \frac{\sqrt{3}}{1} = \sqrt{3}

So the principal argument is

\theta = \frac{\pi}{3}

Thus,

1 + i\sqrt{3} = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)

Similarly,

1 - i\sqrt{3} = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)

Now apply De Moivre's theorem:

(1 + i\sqrt{3})^{2n} = 2^{2n}\left(\cos\frac{2n\pi}{3} + i\sin\frac{2n\pi}{3}\right)

and

(1 - i\sqrt{3})^{2n} = 2^{2n}\left(\cos\left(-\frac{2n\pi}{3}\right) + i\sin\left(-\frac{2n\pi}{3}\right)\right)

Adding these two expressions gives

2^{2n}\left[\cos\frac{2n\pi}{3} + i\sin\frac{2n\pi}{3} + \cos\left(-\frac{2n\pi}{3}\right) + i\sin\left(-\frac{2n\pi}{3}\right)\right]

Using

\cos(-x) = \cos x

and

\sin(-x) = -\sin x

, the imaginary parts cancel, so we get

2^{2n}\left[2\cos\frac{2n\pi}{3}\right]

Therefore,

(1 + i\sqrt{3})^{2n} + (1 - i\sqrt{3})^{2n} = 2^{2n+1}\cos\frac{2n\pi}{3}

Answer:

The correct option is A:

2^{2n+1}\cos\left(\frac{2n\pi}{3}\right)

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