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 \(\frac{1+i}{1-i}\)

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

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

### Step 2: Calculate the numerator and denominator

Calculating the numerator:

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

Calculating the denominator:

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

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: Determine when \(i^{x} = 1\)

The powers of \(i\) cycle every 4:

- \(i^0 = 1\)
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\) (and repeats)

Thus, \(i^{x} = 1\) when \(x\) is a multiple of 4:

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

### Conclusion

The solution to the equation is:

\[ x = 4n \quad (n \in \mathbb{N}) \]

Thus, the correct answer is (B) \(x = 4n\).

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