Question

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

• A. \(2^{2n+1}cos(\frac{2nπ}{3})\)
• B. \(2^{2n}cos(\frac{nπ}{3})\)
• C. \(2^{2n+1}cos(\frac{nπ}{3})\)
• D. \(2^ncos(\frac{2nπ}{3})\)

Hint:

Write each number in polar form, apply De Moivre, and add the conjugate terms.

Answer 1

The Correct Option is A

Step 1: Convert to polar form
For \(1+i\sqrt3\) the modulus is \(\sqrt{1+3} = 2\) and the argument is \(\tan^{-1}\sqrt3 = \frac{\pi}{3}\). So \(1+i\sqrt3 = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\). Its conjugate is \(1-i\sqrt3 = 2\left(\cos\frac{\pi}{3} - i\sin\frac{\pi}{3}\right)\).

Step 2: Apply De Moivre
\[ (1\pm i\sqrt3)^{2n} = 2^{2n}\left(\cos\frac{2n\pi}{3} \pm i\sin\frac{2n\pi}{3}\right) \]

Step 3: Add the two terms
The sine parts cancel, and the cosine parts double:
\[ 2^{2n}\cdot 2\cos\frac{2n\pi}{3} = 2^{2n+1}\cos\frac{2n\pi}{3} \]

Step 4: Check the options
Option (A) matches. Option (C) has the right power of 2 but the wrong angle \(\frac{n\pi}{3}\). Options (B) and (D) have the wrong power of 2. Test \(n=1\): \((1+i\sqrt3)^2 = -2+2i\sqrt3\), and the sum with its conjugate is \(-4\). Option (A) gives \(2^3\cos\frac{2\pi}{3} = 8(-\frac12) = -4\), which agrees.

Final Answer:
The expression equals 2^(2n+1) cos(2n pi / 3). \[ \boxed{\text{(A)}\ 2^{2n+1}\cos\left(\frac{2n\pi}{3}\right)} \]