Find the modulus of `(1-i)^(-2) + (1+ i)^(-2)`

Answer

To find the modulus of \((1-i)^{-2} + (1+i)^{-2}\), we will follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression:
\[ z = (1-i)^{-2} + (1+i)^{-2} \]
This can be expressed as:
\[ z = \frac{1}{(1-i)^2} + \frac{1}{(1+i)^2} \]

### Step 2: Calculate \((1-i)^2\) and \((1+i)^2\)
Next, we calculate \((1-i)^2\) and \((1+i)^2\):
\[ (1-i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i \]
\[ (1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i \]

### Step 3: Substitute back into the expression
Now we substitute these results back into our expression for \(z\):
\[ z = \frac{1}{-2i} + \frac{1}{2i} \]

### Step 4: Simplify the fractions
We simplify each term:
\[ \frac{1}{-2i} = -\frac{1}{2i} \quad \text{and} \quad \frac{1}{2i} = \frac{1}{2i} \]
Thus,
\[ z = -\frac{1}{2i} + \frac{1}{2i} = 0 \]

### Step 5: Find the modulus
The modulus of \(z\) is:
\[ |z| = |0| = 0 \]

### Final Answer
Therefore, the modulus of \((1-i)^{-2} + (1+i)^{-2}\) is:
\[ \boxed{0} \]

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