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

Options

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

Explanation

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

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

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

Using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), where \( a = 1 \) and \( b = i \):

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

Calculating each term:

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

Putting it all together:

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

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

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

### Step 3: Rationalize the denominator
To rationalize the denominator, we multiply the numerator and the denominator by \( -i \):

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

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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