If `((1 +i)/(1 -i))^(x) =1`, then (A) x=2n+1 (B) x=4n (C) x=2n (D) x=4n+1, n` in`N.

Options

A. `x = 2n + 1`
B. `x = 4n`
C. `x = 2n`
D. `x = 4n + 1`

Explanation

To solve the equation \(\left(\frac{1 + i}{1 - i}\right)^{x} = 1\), we will follow these steps:

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

We can simplify this expression by multiplying the numerator and the 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:

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

Thus, we have:

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

### Step 3: Substitute back into the equation

Now we substitute back into the original equation:

\[ (i)^{x} = 1 \]

### Step 4: Analyze the powers of \(i\)

We know that:

\[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1 \]

The powers of \(i\) repeat every 4 terms. Therefore, \(i^x = 1\) when \(x\) is a multiple of 4. This can be expressed as:

\[ x = 4n \quad \text{where } n \in \mathbb{N} \]

### Conclusion

Thus, the solution to the equation is:

\[ \boxed{x = 4n} \]