If `z_(1)=2-i` and `z_(2)=-2+i`, then `Re((1)/(z_(1)overline(z_(2))))=` . . . .

Answer

To solve the problem, we need to find the real part of the expression \(\frac{1}{z_1 \overline{z_2}}\), where \(z_1 = 2 - i\) and \(z_2 = -2 + i\).

### Step-by-Step Solution:

1. **Find the conjugate of \(z_2\)**:
The conjugate of a complex number \(a + bi\) is \(a - bi\). Therefore, for \(z_2 = -2 + i\):
\[ \overline{z_2} = -2 - i \]

2. **Multiply \(z_1\) and \(\overline{z_2}\)**:
Now we need to calculate \(z_1 \overline{z_2}\):
\[ z_1 \overline{z_2} = (2 - i)(-2 - i) \]
Using the distributive property (FOIL):
\[ = 2 \cdot (-2) + 2 \cdot (-i) - i \cdot (-2) - i \cdot (-i) \]
\[ = -4 - 2i + 2i - i^2 \]
Since \(i^2 = -1\), we have:
\[ -i^2 = 1 \]
Thus, the expression simplifies to:
\[ = -4 + 1 = -3 \]

3. **Calculate \(\frac{1}{z_1 \overline{z_2}}\)**:
Now we need to find:
\[ \frac{1}{z_1 \overline{z_2}} = \frac{1}{-3} \]

4. **Find the real part**:
The real part of \(\frac{1}{-3}\) is simply:
\[ \text{Re}\left(\frac{1}{-3}\right) = -\frac{1}{3} \]

### Final Answer:
\[ \text{Re}\left(\frac{1}{z_1 \overline{z_2}}\right) = -\frac{1}{3} \]