If `((1+i)/(1-i))^(x)=1`, then

Options

A. A)x= `2n+1`
B. B)`x=4n`
C. C)`x=2n`
D. D)`x=4n+1`,where `n in N `

Answer

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

### Step 1: Understand the equation
The equation states that the expression \(\left(\frac{1+i}{1-i}\right)^{x}\) is equal to 1. We know that any complex number raised to the power of \(x\) equals 1 if the exponent is a multiple of \(4\) (since \(e^{i\theta}\) returns to 1 for \(\theta = 2\pi n\) where \(n\) is an integer).

### Step 2: Simplify the fraction
We can simplify \(\frac{1+i}{1-i}\) by multiplying 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 3: Calculate the numerator and denominator
Calculating the numerator:

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

Calculating the denominator:

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

Thus, we have:

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

### Step 4: Substitute back into the equation
Now substituting back into the original equation gives us:

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

### Step 5: Express 1 in terms of powers of \(i\)
We know that \(1\) can be expressed as \(i^{4n}\) for any integer \(n\) (since \(i^4 = 1\)). Therefore, we can write:

\[ i^{x} = i^{4n} \]

### Step 6: Compare the exponents
Since the bases are the same, we can equate the exponents:

\[ x = 4n \]

where \(n\) is a natural number.

### Conclusion
Thus, the solution to the equation is:

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