If z and w are two non - zero complex numbers such that `|zw|=1 and arg(z)-arg(w)=(pi)/(2),` then the value of `5ibarzw` is equal to

Options

A. `-5`
B. 5i
C. 5
D. `-5i`

Answer

To solve the problem, we need to find the value of \(5i \bar{z} w\) given that \( |zw| = 1 \) and \( \arg(z) - \arg(w) = \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Understanding the Conditions**:
- We know that \( |zw| = 1 \). This implies that \( |z| \cdot |w| = 1 \).
- Let \( |z| = r_1 \) and \( |w| = r_2 \). Therefore, \( r_1 \cdot r_2 = 1 \).

2. **Expressing the Arguments**:
- The condition \( \arg(z) - \arg(w) = \frac{\pi}{2} \) can be rewritten as:
\[ \arg(z) = \arg(w) + \frac{\pi}{2} \]
- Let \( \arg(w) = \theta \). Then, we can express \( z \) and \( w \) as:
\[ w = r_2 e^{i\theta} \]
\[ z = r_1 e^{i(\theta + \frac{\pi}{2})} = r_1 e^{i\theta} e^{i\frac{\pi}{2}} = r_1 e^{i\theta} i \]

3. **Finding the Conjugate of \( z \)**:
- The conjugate of \( z \) is:
\[ \bar{z} = \overline{r_1 e^{i(\theta + \frac{\pi}{2})}} = r_1 e^{-i(\theta + \frac{\pi}{2})} = r_1 e^{-i\theta} e^{-i\frac{\pi}{2}} = r_1 e^{-i\theta} (-i) \]

4. **Calculating \( 5i \bar{z} w \)**:
- Now we can substitute \( \bar{z} \) and \( w \) into the expression:
\[ 5i \bar{z} w = 5i \left( r_1 e^{-i\theta} (-i) \right) \left( r_2 e^{i\theta} \right) \]
- Simplifying this:
\[ = 5i (-i) r_1 r_2 e^{-i\theta} e^{i\theta} = 5(-1) r_1 r_2 \]
- Since \( r_1 r_2 = 1 \):
\[ = 5(-1) = -5 \]

5. **Final Result**:
- Thus, the value of \( 5i \bar{z} w \) is:
\[ \boxed{-5} \]