Let `z_(1)=2-i` and `z_(2)=2+i`, then `"Im"((1)/(z_(1)z_(2)))` is

Options

A. 0
B. 6
C. 2
D. 8

Answer

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

### Step-by-Step Solution:

1. **Identify the values of \(z_1\) and \(z_2\)**:
\[ z_1 = 2 - i, \quad z_2 = 2 + i \]

2. **Calculate the product \(z_1 z_2\)**:
\[ z_1 z_2 = (2 - i)(2 + i) \]
Using the formula for the difference of squares, \(a^2 - b^2\):
\[ = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5 \]

3. **Calculate \(\frac{1}{z_1 z_2}\)**:
\[ \frac{1}{z_1 z_2} = \frac{1}{5} \]

4. **Express \(\frac{1}{5}\) in the form \(a + ib\)**:
\[ \frac{1}{5} = \frac{1}{5} + i \cdot 0 \]
Here, \(a = \frac{1}{5}\) and \(b = 0\).

5. **Identify the imaginary part**:
The imaginary part of \(\frac{1}{z_1 z_2}\) is \(b\):
\[ \text{Im}\left(\frac{1}{z_1 z_2}\right) = 0 \]

### Final Answer:
The imaginary part of \(\frac{1}{z_1 z_2}\) is \(0\).