Question

If \( n, K \in \mathbb{N} \) such that \( n \neq 3K \), then \( (\sqrt{3}+i)^{2n} + (\sqrt{3}-i)^{2n} = \)

• A. \( (-1)^n 2^{2n+1} \)
• B. \( (-1)^{n+1} 2^{2n+1} \)
• C. \( (-1)^{n+1} 2^{2n} \)
• D. \( (-1)^{n+1} 2^n \)

Hint:

Checking specific small values for \( n \) (like \( n=1 \) and \( n=2 \)) is a fast way to verify the formula from the options in competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

Convert the complex numbers to polar form to apply exponentiation easily.
Step 2: Detailed Explanation:

Let \( z = \sqrt{3} + i \). Modulus \( r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \). Argument \( \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \). So, \( z = 2e^{i\pi/6} \). Its conjugate is \( \bar{z} = \sqrt{3} - i = 2e^{-i\pi/6} \). The expression is: \[ E = z^{2n} + \bar{z}^{2n} \] \[ E = (2e^{i\pi/6})^{2n} + (2e^{-i\pi/6})^{2n} \] \[ E = 2^{2n} e^{in\pi/3} + 2^{2n} e^{-in\pi/3} \] \[ E = 2^{2n} (e^{in\pi/3} + e^{-in\pi/3}) \] Using \( e^{ix} + e^{-ix} = 2\cos x \): \[ E = 2^{2n} \cdot 2\cos\left(\frac{n\pi}{3}\right) = 2^{2n+1} \cos\left(\frac{n\pi}{3}\right) \] Now analyze the term \( \cos\left(\frac{n\pi}{3}\right) \) given that \( n \) is not a multiple of 3 (\( n \neq 3K \)). The possible values for \( n \pmod 6 \) excluding multiples of 3 (0 and 3) are: 1. \( n \equiv 1, 5 \pmod 6 \) (odd \( n \)): \( \cos(\pi/3) = \frac{1}{2} \), \( \cos(5\pi/3) = \frac{1}{2} \). Here \( n \) is odd, so \( n+1 \) is even, meaning \( (-1)^{n+1} = 1 \). Result: \( 2^{2n+1}(\frac{1}{2}) = 2^{2n} \). This matches \( (-1)^{n+1} 2^{2n} \). 2. \( n \equiv 2, 4 \pmod 6 \) (even \( n \)): \( \cos(2\pi/3) = -\frac{1}{2} \), \( \cos(4\pi/3) = -\frac{1}{2} \). Here \( n \) is even, so \( n+1 \) is odd, meaning \( (-1)^{n+1} = -1 \). Result: \( 2^{2n+1}(-\frac{1}{2}) = -2^{2n} \). This also matches \( (-1)^{n+1} 2^{2n} \). Thus, in both cases, the expression simplifies to \( (-1)^{n+1} 2^{2n} \).
Step 4: Final Answer:

The value is \( (-1)^{n+1} 2^{2n} \).