Let `z_1=2-i ,z_2=-2+i`. Find
(i) Re `((z_1z_2)/( bar z_1))`
(ii) Im`(1/(z_1 bar z_1))`

Options

A. `2/5`
B. `-2/5`
C. `5/2`
D. `-2`

Explanation

To solve the given problem, we will break it down into two parts as specified in the question.

### Given:
- \( z_1 = 2 - i \)
- \( z_2 = -2 + i \)

### Part (i): Find \( \text{Re} \left( \frac{z_1 z_2}{\overline{z_1}} \right) \)

1. **Calculate \( z_1 z_2 \)**:
\[ z_1 z_2 = (2 - i)(-2 + i) \]
Using the distributive property (FOIL):
\[ = 2 \cdot (-2) + 2 \cdot i - i \cdot (-2) - i \cdot i \]
\[ = -4 + 2i + 2i - i^2 \]
Since \( i^2 = -1 \):
\[ = -4 + 4i + 1 = -3 + 4i \]

2. **Calculate \( \overline{z_1} \)**:
\[ \overline{z_1} = 2 + i \]

3. **Now compute \( \frac{z_1 z_2}{\overline{z_1}} \)**:
\[ \frac{z_1 z_2}{\overline{z_1}} = \frac{-3 + 4i}{2 + i} \]

4. **Multiply numerator and denominator by the conjugate of the denominator**:
\[ = \frac{(-3 + 4i)(2 - i)}{(2 + i)(2 - i)} \]

5. **Calculate the denominator**:
\[ (2 + i)(2 - i) = 4 - i^2 = 4 + 1 = 5 \]

6. **Calculate the numerator**:
\[ (-3 + 4i)(2 - i) = -6 + 3i + 8i - 4i^2 \]
\[ = -6 + 11i + 4 = -2 + 11i \]

7. **Now combine the results**:
\[ \frac{-2 + 11i}{5} = -\frac{2}{5} + \frac{11}{5}i \]

8. **Find the real part**:
\[ \text{Re} \left( \frac{z_1 z_2}{\overline{z_1}} \right) = -\frac{2}{5} \]

### Part (ii): Find \( \text{Im} \left( \frac{1}{z_1 \overline{z_1}} \right) \)

1. **Calculate \( z_1 \overline{z_1} \)**:
\[ z_1 \overline{z_1} = (2 - i)(2 + i) = 4 + 1 = 5 \]

2. **Now compute \( \frac{1}{z_1 \overline{z_1}} \)**:
\[ \frac{1}{z_1 \overline{z_1}} = \frac{1}{5} \]

3. **Express \( \frac{1}{5} \) in terms of real and imaginary parts**:
\[ \frac{1}{5} = \frac{1}{5} + 0i \]

4. **Find the imaginary part**:
\[ \text{Im} \left( \frac{1}{z_1 \overline{z_1}} \right) = 0 \]

### Final Answers:
(i) \( \text{Re} \left( \frac{z_1 z_2}{\overline{z_1}} \right) = -\frac{2}{5} \)

(ii) \( \text{Im} \left( \frac{1}{z_1 \overline{z_1}} \right) = 0 \)

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