The smallest positive integral value of n for which `(1+sqrt3i)^(n/2)` is real is

Options

A. 3
B. 6
C. 12
D. 0

Explanation

To find the smallest positive integral value of \( n \) for which \( (1 + \sqrt{3}i)^{n/2} \) is real, we can follow these steps:

### Step 1: Convert the complex number to polar form
The complex number \( 1 + \sqrt{3}i \) can be expressed in polar form. First, we find the modulus \( r \) and the argument \( \theta \).

\[ r = |1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Next, we find the argument \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

Thus, we can express \( 1 + \sqrt{3}i \) in polar form as:
\[ 1 + \sqrt{3}i = 2 \left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) \]

### Step 2: Raise to the power \( n/2 \)
Now we raise this expression to the power \( n/2 \):
\[ (1 + \sqrt{3}i)^{n/2} = \left(2 \left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\right)^{n/2} = 2^{n/2} \left(\cos\left(\frac{n\pi}{6}\right) + i\sin\left(\frac{n\pi}{6}\right)\right) \]

### Step 3: Determine when the expression is real
For the expression \( (1 + \sqrt{3}i)^{n/2} \) to be real, the imaginary part must be zero. This occurs when:
\[ \sin\left(\frac{n\pi}{6}\right) = 0 \]

### Step 4: Solve for \( n \)
The sine function is zero at integer multiples of \( \pi \):
\[ \frac{n\pi}{6} = k\pi \quad \text{for } k \in \mathbb{Z} \]
This simplifies to:
\[ n = 6k \]

The smallest positive integral value of \( n \) occurs when \( k = 1 \):
\[ n = 6 \times 1 = 6 \]

### Conclusion
Thus, the smallest positive integral value of \( n \) for which \( (1 + \sqrt{3}i)^{n/2} \) is real is:
\[ \boxed{6} \]