The least positive integer n for which `((1+i sqrt3)/(1-isqrt3))^(n)=1,` is

Options

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

Answer

To solve the problem, we need to find the least positive integer \( n \) such that

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

### Step 1: Simplify the expression

First, we simplify the expression inside the parentheses:

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

To do this, we multiply the numerator and the denominator by the conjugate of the denominator:

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

### Step 2: Calculate the denominator

The denominator can be simplified as follows:

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

### Step 3: Calculate the numerator

Now, we calculate the numerator:

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

### Step 4: Combine the results

Now we can combine the results:

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

### Step 5: Rewrite the expression

Thus, we have:

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

### Step 6: Convert to polar form

Next, we convert \(\frac{-1 + i\sqrt{3}}{2}\) to polar form. The modulus \( r \) is given by:

\[ r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1. \]

The argument \( \theta \) can be calculated as:

\[ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) = \tan^{-1}(-\sqrt{3}). \]

This angle corresponds to \( \frac{2\pi}{3} \) in the second quadrant.

### Step 7: Write in polar form

Thus, we can write:

\[ \frac{-1 + i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}. \]

### Step 8: Raise to the power of \( n \)

Now we have:

\[ \left(e^{i\frac{2\pi}{3}}\right)^n = e^{i2\pi k} \quad \text{for some integer } k. \]

This implies:

\[ n \cdot \frac{2\pi}{3} = 2\pi k. \]

### Step 9: Solve for \( n \)

Dividing both sides by \( 2\pi \):

\[ \frac{n}{3} = k \implies n = 3k. \]

The smallest positive integer \( n \) occurs when \( k = 1 \):

\[ n = 3. \]

### Final Answer

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

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

is

\[ \boxed{3}. \]