`((1+i)/(1-i))^(2) + ((1-i)/(1+i))^(2)` is equal to :

Options

A. `2i`
B. `-2i`
C. `-2`
D. 2

Answer

To solve the expression \(\left(\frac{1+i}{1-i}\right)^{2} + \left(\frac{1-i}{1+i}\right)^{2}\), we will follow these steps:

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

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

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

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 expression

Now we substitute back into the expression:

\[ \left(\frac{1+i}{1-i}\right)^{2} + \left(\frac{1-i}{1+i}\right)^{2} = (i)^{2} + (-i)^{2} \]

Calculating \(i^{2}\) and \((-i)^{2}\):

\[ i^{2} = -1 \quad \text{and} \quad (-i)^{2} = (-1)^{2} \cdot i^{2} = 1 \cdot (-1) = -1 \]

Thus, we have:

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

### Final Answer

The value of the expression is \(-2\).

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