If `(sqrt(3)-i)^n=2^n, n in I`, the set of integers, then n is a multiple of

Options

A. 6
B. 10
C. 9
D. 12

Explanation

To solve the problem, we need to determine the values of \( n \) such that \( (\sqrt{3} - i)^n = 2^n \), where \( n \) is an integer.

### Step-by-Step Solution:

1. **Express the complex number in polar form**:
The complex number \( \sqrt{3} - i \) can be expressed in polar form. We first find its modulus:
\[ r = | \sqrt{3} - i | = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2. \]
Next, we find the argument (angle) \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}. \]
Therefore, we can write:
\[ \sqrt{3} - i = 2 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right). \]

2. **Raise to the power of \( n \)**:
Now, we raise this expression to the power of \( n \):
\[ (\sqrt{3} - i)^n = \left(2 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right)\right)\right)^n = 2^n \left( \cos\left(-\frac{n\pi}{6}\right) + i \sin\left(-\frac{n\pi}{6}\right) \right). \]

3. **Set the equation**:
We have:
\[ 2^n \left( \cos\left(-\frac{n\pi}{6}\right) + i \sin\left(-\frac{n\pi}{6}\right) \right) = 2^n. \]
Dividing both sides by \( 2^n \) (assuming \( n \neq 0 \)):
\[ \cos\left(-\frac{n\pi}{6}\right) + i \sin\left(-\frac{n\pi}{6}\right) = 1. \]

4. **Equate real and imaginary parts**:
From the equation above, we can equate the real and imaginary parts:
- Real part: \( \cos\left(-\frac{n\pi}{6}\right) = 1 \)
- Imaginary part: \( \sin\left(-\frac{n\pi}{6}\right) = 0 \)

5. **Solve for \( n \)**:
- The equation \( \cos\left(-\frac{n\pi}{6}\right) = 1 \) implies:
\[ -\frac{n\pi}{6} = 2k\pi \quad \text{for some integer } k \implies n = -12k. \]
- The equation \( \sin\left(-\frac{n\pi}{6}\right) = 0 \) implies:
\[ -\frac{n\pi}{6} = m\pi \quad \text{for some integer } m \implies n = -6m. \]

6. **Find common multiples**:
The values of \( n \) must satisfy both conditions. The least common multiple of \( 12 \) and \( 6 \) is \( 12 \). Thus, \( n \) must be a multiple of \( 12 \).

### Final Answer:
Thus, \( n \) is a multiple of \( 12 \).