If `((1+i)/(1-i))^(3) - (( 1-i)/( 1+i))^(3) = x+iy` , then (x,y) is equal to

Options

A. A)( 0,-2)
B. B)(-2,0)
C. C)(0,2)
D. D)(2,0)

Answer

To solve the equation \(\left(\frac{1+i}{1-i}\right)^{3} - \left(\frac{1-i}{1+i}\right)^{3} = x + iy\), we will follow these steps:

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

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

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

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 2: Simplify \(\frac{1-i}{1+i}\)

Similarly, we simplify \(\frac{1-i}{1+i}\) by multiplying by the conjugate of the denominator:

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

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 3: Substitute back into the original equation

Now we substitute these results back into the original equation:

\[ \left(i\right)^{3} - \left(-i\right)^{3} \]

Calculating \(i^{3}\):

\[ i^{3} = i^{2} \cdot i = -1 \cdot i = -i \]

Calculating \((-i)^{3}\):

\[ (-i)^{3} = -i^{3} = -(-i) = i \]

Thus, we have:

\[ -i - i = -2i \]

### Step 4: Compare with \(x + iy\)

Now we compare \(-2i\) with \(x + iy\):

\[ x + iy = 0 - 2i \]

From this comparison, we can identify:

\[ x = 0 \quad \text{and} \quad y = -2 \]

### Final Answer

Thus, the values of \(x\) and \(y\) are:

\[ (x, y) = (0, -2) \]