Let `z_(1)=2 -I, z_(2)= -2 +i`, find (i) Re `((z_(1)z_(2))/(bar(z)_(1)))`, (ii) Im `((1)/(z_(1)bar(z)_(2)))`

Answer

To solve the problem, we will break it down into two parts as given 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 \( \overline{z_1} \)**:
\[ \overline{z_1} = \overline{(2 - i)} = 2 + i \]

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

3. **Now, substitute into the expression**:
\[ \frac{z_1 z_2}{\overline{z_1}} = \frac{-3 + 4i}{2 + i} \]

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

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

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

7. **Putting it all together**:
\[ \frac{-2 + 11i}{5} = -\frac{2}{5} + \frac{11}{5}i \]

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

### Part (ii): Find \( \text{Im} \left( \frac{1}{z_1 \overline{z_2}} \right) \)

1. **Calculate \( \overline{z_2} \)**:
\[ \overline{z_2} = \overline{(-2 + i)} = -2 - i \]

2. **Calculate \( z_1 \overline{z_2} \)**:
\[ z_1 \overline{z_2} = (2 - i)(-2 - i) \]
Using the distributive property:
\[ = 2 \cdot (-2) + 2 \cdot (-i) - i \cdot (-2) - i \cdot (-i) \]
\[ = -4 - 2i + 2i - 1 = -4 - 1 = -5 \]

3. **Now, substitute into the expression**:
\[ \frac{1}{z_1 \overline{z_2}} = \frac{1}{-5} = -\frac{1}{5} \]

4. **Identify the imaginary part**:
\[ \text{Im} \left( \frac{1}{z_1 \overline{z_2}} \right) = 0 \]

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

(ii) \( \text{Im} \left( \frac{1}{z_1 \overline{z_2}} \right) = 0 \)