The least positive integeral value of n for which `(sqrt(3)+i)^(n)=(sqrt(3)-i)^(n)`, is

Options

A. 3
B. 4
C. 6
D. none of these

Explanation

To find the least positive integral value of \( n \) for which \( (\sqrt{3} + i)^n = (\sqrt{3} - i)^n \), we can follow these steps:

### Step 1: Set up the equation
We start with the equation:
\[ (\sqrt{3} + i)^n = (\sqrt{3} - i)^n \]

### Step 2: Divide both sides
Dividing both sides by \( (\sqrt{3} - i)^n \), we get:
\[ \frac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^n} = 1 \]
This simplifies to:
\[ \left( \frac{\sqrt{3} + i}{\sqrt{3} - i} \right)^n = 1 \]

### Step 3: Simplify the fraction
To simplify \( \frac{\sqrt{3} + i}{\sqrt{3} - i} \), we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(\sqrt{3} + i)(\sqrt{3} + i)}{(\sqrt{3} - i)(\sqrt{3} + i)} = \frac{(\sqrt{3} + i)^2}{(\sqrt{3})^2 - (i)^2} \]
Calculating the denominator:
\[ (\sqrt{3})^2 - (i)^2 = 3 - (-1) = 4 \]
Calculating the numerator:
\[ (\sqrt{3} + i)^2 = 3 + 2\sqrt{3}i - 1 = 2 + 2\sqrt{3}i \]
Thus, we have:
\[ \frac{2 + 2\sqrt{3}i}{4} = \frac{1 + \sqrt{3}i}{2} \]

### Step 4: Express in polar form
Next, we express \( \frac{1 + \sqrt{3}i}{2} \) in polar form. The modulus is:
\[ \left| \frac{1 + \sqrt{3}i}{2} \right| = \frac{1}{2} \sqrt{1^2 + (\sqrt{3})^2} = \frac{1}{2} \sqrt{4} = 1 \]
The argument \( \theta \) is given by:
\[ \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]
Thus, we can write:
\[ \frac{1 + \sqrt{3}i}{2} = e^{i\frac{\pi}{3}} \]

### Step 5: Substitute back into the equation
Now substituting back, we have:
\[ \left( e^{i\frac{\pi}{3}} \right)^n = 1 \]
This implies:
\[ e^{i\frac{n\pi}{3}} = e^{i2k\pi} \quad \text{for some integer } k \]

### Step 6: Solve for \( n \)
From this, we can equate the exponents:
\[ \frac{n\pi}{3} = 2k\pi \implies n = 6k \]
The least positive integer value for \( n \) occurs when \( k = 1 \):
\[ n = 6 \]

### Conclusion
Thus, the least positive integral value of \( n \) is:
\[ \boxed{6} \]