If `z=1+i,` then what is the inverse of `z^(2)`?

Options

A. `1-i`
B. `i//2`
C. `-i//2`
D. `2i`

Answer

To find the inverse of \( z^2 \) where \( z = 1 + i \), we can follow these steps:

### Step 1: Calculate \( z^2 \)
Given \( z = 1 + i \), we need to calculate \( z^2 \).

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

### Step 2: Expand using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \)
Using the expansion formula:

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

### Step 3: Substitute the values
Calculating the individual terms:

- \( 1^2 = 1 \)
- \( 2(1)(i) = 2i \)
- \( i^2 = -1 \)

So, substituting these values into the equation:

\[ z^2 = 1 + 2i - 1 \]

### Step 4: Simplify the expression
Now, simplify the expression:

\[ z^2 = 2i \]

### Step 5: Find the inverse of \( z^2 \)
The inverse of \( z^2 \) is given by:

\[ \frac{1}{z^2} = \frac{1}{2i} \]

### Step 6: Rationalize the denominator
To rationalize the denominator, multiply the numerator and denominator by \( i \):

\[ \frac{1}{2i} \cdot \frac{i}{i} = \frac{i}{2i^2} \]

### Step 7: Substitute \( i^2 \) with \(-1\)
Since \( i^2 = -1 \):

\[ \frac{i}{2(-1)} = \frac{i}{-2} = -\frac{i}{2} \]

### Final Answer
Thus, the inverse of \( z^2 \) is:

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

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