If `((1 - i)/(1+ i))^(100) = x + iy` then the value of `(x, y)` is:

Options

A. `(0,1)`
B. `(0,0)`
C. `(1,0)`
D. `(-1,0)`

Answer

To solve the problem \( \left(\frac{1 - i}{1 + i}\right)^{100} = x + iy \), we will follow these steps:

### Step 1: Rationalize the denominator
We start by rationalizing the expression \( \frac{1 - i}{1 + i} \). To do this, we multiply the numerator and 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: Simplify the denominator
Now, we simplify the denominator:

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

### Step 3: Simplify the numerator
Next, we simplify the numerator:

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

### Step 4: Combine the results
Now we can combine the results:

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

### Step 5: Raise to the power of 100
Next, we raise \( -i \) to the power of 100:

\[ (-i)^{100} \]

### Step 6: Use properties of powers of \( i \)
Recall that \( i^2 = -1 \), and thus \( (-i)^2 = -1 \). Therefore, we can express \( (-i)^{100} \) as:

\[ (-i)^{100} = (i^2)^{50} = (-1)^{50} = 1 \]

### Step 7: Write in the form \( x + iy \)
Now we have:

\[ 1 = x + iy \]

### Step 8: Identify \( x \) and \( y \)
From the equation \( 1 = x + iy \), we can see that:

\[ x = 1 \quad \text{and} \quad y = 0 \]

Thus, the final answer is:

\[ (x, y) = (1, 0) \]