The last positive integer n for which `((1+i)/(1-i))^(n)` is real, is

Options

A. 2
B. 4
C. 8
D. none of these

Explanation

To solve the problem, we need to find the last positive integer \( n \) for which

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

is a real number.

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

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

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

### 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:

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

So, we have:

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

### Step 3: Raise to the power \( n \)

Now, we need to consider:

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

### Step 4: Determine when \( i^n \) is real

The powers of \( i \) cycle through four values:

- \( i^1 = i \) (not real)
- \( i^2 = -1 \) (real)
- \( i^3 = -i \) (not real)
- \( i^4 = 1 \) (real)

This cycle repeats every four integers. Therefore, \( i^n \) is real when \( n \equiv 0 \) or \( 2 \mod 4 \).

### Step 5: Find the last positive integer \( n \)

The last positive integer \( n \) that satisfies this condition is \( n = 2 \) because:

- For \( n = 2 \), \( i^2 = -1 \) (real)
- For \( n = 4 \), \( i^4 = 1 \) (real)
- For \( n = 6 \), \( i^6 = -1 \) (real)
- For \( n = 8 \), \( i^8 = 1 \) (real)

However, since we are looking for the last positive integer \( n \) before it becomes non-real, we find that the last positive integer \( n \) for which \( i^n \) is real is indeed \( n = 2 \).

### Final Answer

Thus, the last positive integer \( n \) for which \(\left(\frac{1+i}{1-i}\right)^n\) is real is:

\[ \boxed{2} \]