If `z` and `omega` are two non-zero complex numbers such that `|zomega|=1` and `arg(z)-arg(omega)=pi/2`, then `barzomega` is equal to

Options

A. `-i`
B. 1
C. `-1`
D. i

Explanation

To solve the problem step by step, we start with the given conditions and manipulate them to find the value of \( \overline{z} \omega \).

### Step 1: Understand the given conditions
We have two non-zero complex numbers \( z \) and \( \omega \) such that:
1. \( |z \omega| = 1 \)
2. \( \arg(z) - \arg(\omega) = \frac{\pi}{2} \)

### Step 2: Express the argument condition
From the second condition, we can express it as:
\[ \arg(z) = \arg(\omega) + \frac{\pi}{2} \]
This means that \( z \) can be expressed in terms of \( \omega \):
\[ \frac{z}{\omega} = e^{i \frac{\pi}{2}} = i \]
Thus, we can write:
\[ z = i \omega \]

### Step 3: Use the modulus condition
From the first condition \( |z \omega| = 1 \):
\[ |z| \cdot |\omega| = 1 \]
Substituting \( z = i \omega \):
\[ |i \omega| \cdot |\omega| = 1 \]
Since \( |i| = 1 \), we have:
\[ |\omega|^2 = 1 \implies |\omega| = 1 \]
Thus, \( |z| = \frac{1}{|\omega|} = 1 \).

### Step 4: Find \( \overline{z} \omega \)
We know that:
\[ \overline{z} = \overline{i \omega} = -i \overline{\omega} \]
Now, we can find \( \overline{z} \omega \):
\[ \overline{z} \omega = (-i \overline{\omega}) \omega = -i |\omega|^2 \]
Since \( |\omega| = 1 \):
\[ \overline{z} \omega = -i \]

### Final Answer
Thus, the value of \( \overline{z} \omega \) is:
\[ \overline{z} \omega = -i \]