The least positive integer n for which `((1+sqrt(3)i)/(1-sqrt(3)i))^(n) = 1`, is

Options

A. 2
B. 3
C. 5
D. 6

Answer

To solve the problem of finding the least positive integer \( n \) such that

\[ \left(\frac{1 + \sqrt{3}i}{1 - \sqrt{3}i}\right)^{n} = 1, \]

we can follow these steps:

### Step 1: Simplify the expression
We start by simplifying the expression \( \frac{1 + \sqrt{3}i}{1 - \sqrt{3}i} \). To do this, we multiply the numerator and the denominator by the conjugate of the denominator:

\[ \frac{(1 + \sqrt{3}i)(1 + \sqrt{3}i)}{(1 - \sqrt{3}i)(1 + \sqrt{3}i)}. \]

### Step 2: Calculate the denominator
The denominator simplifies as follows:

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

### Step 3: Calculate the numerator
Now, we calculate the numerator:

\[ (1 + \sqrt{3}i)(1 + \sqrt{3}i) = 1 + 2\sqrt{3}i + 3i^2 = 1 + 2\sqrt{3}i - 3 = -2 + 2\sqrt{3}i. \]

### Step 4: Combine results
Now we can write the simplified form:

\[ \frac{-2 + 2\sqrt{3}i}{4} = \frac{-1 + \frac{\sqrt{3}}{2}i}{2} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i. \]

### Step 5: Recognize the complex number
The expression \( -\frac{1}{2} + \frac{\sqrt{3}}{2}i \) can be recognized as \( \omega \), where \( \omega = e^{i\frac{2\pi}{3}} \) (a cube root of unity).

### Step 6: Set up the equation
Thus, we have:

\[ \frac{1 + \sqrt{3}i}{1 - \sqrt{3}i} = \omega. \]

We need to find \( n \) such that:

\[ \omega^n = 1. \]

### Step 7: Find the least positive integer \( n \)
The cube roots of unity satisfy \( \omega^3 = 1 \). Therefore, the least positive integer \( n \) for which \( \omega^n = 1 \) is:

\[ n = 3. \]

### Final Answer
Thus, the least positive integer \( n \) for which

\[ \left(\frac{1 + \sqrt{3}i}{1 - \sqrt{3}i}\right)^{n} = 1 \]

is

\[ \boxed{3}. \]