Let `z_1=2-i ,z_2=-2+i`. Find
Re `((z_1z_2)/( bar z'_1))`

Options

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

Explanation

To solve the problem, we need to find the real part of the expression \(\frac{z_1 z_2}{\bar{z_1}}\), where \(z_1 = 2 - i\) and \(z_2 = -2 + i\).

### Step-by-Step Solution:

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

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

3. **Form the expression \(\frac{z_1 z_2}{\bar{z_1}}\)**:
\[ \frac{z_1 z_2}{\bar{z_1}} = \frac{-3 + 4i}{2 + i} \]

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

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

6. **Calculate the numerator**:
\[ (-3 + 4i)(2 - i) = -3 \cdot 2 + (-3)(-i) + 4i \cdot 2 + 4i \cdot (-i) \]
\[ = -6 + 3i + 8i - 4i^2 \]
Again, since \(i^2 = -1\):
\[ = -6 + 11i + 4 = -2 + 11i \]

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

8. **Identify the real part**:
The real part of \(\frac{z_1 z_2}{\bar{z_1}}\) is:
\[ \text{Re}\left(\frac{z_1 z_2}{\bar{z_1}}\right) = \frac{-2}{5} \]

### Final Answer:
\[ \text{Re}\left(\frac{z_1 z_2}{\bar{z_1}}\right) = -\frac{2}{5} \]