What is the least positive integer n for which `((1+i)/(1-i))^(n)=1` ?

Options

A. 16
B. 12
C. 8
D. 4

Explanation

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

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

### Step 1: Simplify \( \frac{1+i}{1-i} \)

We start by rationalizing the denominator:

\[ \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)(1+i)}{(1-i)(1+i)}. \]

### Step 2: Calculate the numerator and denominator

Calculating the numerator:

\[ (1+i)(1+i) = 1^2 + 2i + i^2 = 1 + 2i - 1 = 2i. \]

Calculating the denominator:

\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2. \]

### Step 3: Combine the results

Now we can combine the results:

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

### Step 4: Set up the equation

Now we have:

\[ (i)^n = 1. \]

### Step 5: Determine when \( i^n = 1 \)

The powers of \( i \) are:

- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)

This pattern repeats every 4 powers. Therefore, \( i^n = 1 \) when \( n \) is a multiple of 4.

### Step 6: Find the least positive integer \( n \)

The least positive integer \( n \) that satisfies this condition is:

\[ n = 4. \]

Thus, the answer is:

\[ \boxed{4}. \]