The least 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 least positive integer \( n \) such that \( \left( \frac{1+i}{1-i} \right)^n \) is a real number.

### Step-by-step Solution:

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

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

The denominator simplifies as follows:
\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 \]

The numerator simplifies as follows:
\[ (1+i)(1+i) = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \]

Thus, we have:
\[ \frac{1+i}{1-i} = \frac{2i}{2} = i \]

2. **Raise to the power of n**:
Now, we need to find \( \left( i \right)^n \). The powers of \( i \) are periodic with a cycle of 4:
\[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1 \]
This cycle repeats every 4 powers.

3. **Determine when \( i^n \) is real**:
For \( i^n \) to be real, \( n \) must be such that \( i^n \) equals either \( 1 \) or \( -1 \). This occurs when:
- \( n \equiv 0 \mod 4 \) (for \( i^n = 1 \))
- \( n \equiv 2 \mod 4 \) (for \( i^n = -1 \))

4. **Find the least positive integer n**:
The smallest positive integer \( n \) that satisfies either of these conditions is \( n = 2 \) (since \( 2 \equiv 2 \mod 4 \)).

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

\[ \boxed{2} \]