The smallest positive integer n for which `((1-i)/(1+i))^(n^(2)) = 1`
where `i=sqrt(-1)`, is

Options

A. 2
B. 4
C. 6
D. 8

Answer

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

\[ \left( \frac{1 - i}{1 + i} \right)^{n^2} = 1 \]

where \( i = \sqrt{-1} \).

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

To simplify \( \frac{1 - i}{1 + i} \), we can multiply the numerator and denominator by the conjugate of the denominator, which is \( 1 - i \):

\[ \frac{1 - i}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{(1 - i)^2}{(1 + i)(1 - i)} \]

### Step 2: Calculate the denominator

The denominator can be calculated as follows:

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

### Step 3: Calculate the numerator

Now, calculate the numerator:

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

### Step 4: Combine results

Now we can combine our results:

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

### Step 5: Set up the equation

Now we have:

\[ (-i)^{n^2} = 1 \]

### Step 6: Determine when \( (-i)^{n^2} = 1 \)

The expression \( (-i)^{n^2} \) can be rewritten using the polar form of complex numbers. The complex number \( -i \) can be expressed as:

\[ -i = e^{-i\frac{\pi}{2}} \]

Thus,

\[ (-i)^{n^2} = \left(e^{-i\frac{\pi}{2}}\right)^{n^2} = e^{-i\frac{n^2 \pi}{2}} \]

For this to equal 1, we need:

\[ -\frac{n^2 \pi}{2} = 2k\pi \quad \text{for some integer } k \]

### Step 7: Solve for \( n^2 \)

This simplifies to:

\[ -\frac{n^2}{2} = 2k \implies n^2 = -4k \]

Since \( n^2 \) must be positive, \( k \) must be negative. Let \( k = -m \) where \( m \) is a positive integer:

\[ n^2 = 4m \]

### Step 8: Find the smallest \( n \)

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

\[ n^2 = 4 \implies n = 2 \]

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

\[ \left( \frac{1 - i}{1 + i} \right)^{n^2} = 1 \]

is

\[ \boxed{2} \]