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: Use Euler form
Write $1\pm i\sqrt3 = 2e^{\pm i\pi/3}$.

Step 2: Raise to the power
$(2e^{i\pi/3})^{2n} = 2^{2n}e^{2in\pi/3}$, and similarly with a minus sign in the exponent for the conjugate.

Step 3: Sum
$e^{i\theta}+e^{-i\theta} = 2\cos\theta$, so the sum is $2\cdot 2^{2n}\cos\frac{2n\pi}{3} = 2^{2n+1}\cos\frac{2n\pi}{3}$.

Step 4: Numerical test
For $n=2$: $(1+i\sqrt3)^4 = 16e^{4i\pi/3}$, real part $16\cos240^{\circ} = -8$. The sum of both is $-16$. Formula: $2^5\cos\frac{4\pi}{3} = 32\cdot(-\tfrac12) = -16$. It matches option (A).

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