Let `z in C` be such that `|z| lt 1 `.If `omega =(5+3z)/(5(1-z)` then

Options

A. `4 Im ( omega) gt 5`
B. `5 Re (omega) gt 1`
C. `5 Im(omega) lt 1 `
D. none

Explanation

To solve the problem given, we need to analyze the expression for \( \omega \) and derive some properties based on the condition \( |z| < 1 \).

### Step-by-Step Solution:

1. **Write the expression for \( \omega \)**:
\[ \omega = \frac{5 + 3z}{5(1 - z)} \]

2. **Multiply both sides by \( 5(1 - z) \)** to eliminate the denominator:
\[ 5\omega(1 - z) = 5 + 3z \]

3. **Distribute \( 5\omega \)**:
\[ 5\omega - 5\omega z = 5 + 3z \]

4. **Rearrange the equation to isolate terms involving \( z \)**:
\[ 5\omega - 5 = 3z + 5\omega z \]

5. **Factor out \( z \) on the right side**:
\[ 5\omega - 5 = z(3 + 5\omega) \]

6. **Solve for \( z \)**:
\[ z = \frac{5\omega - 5}{3 + 5\omega} \]

7. **Substitute \( z \) back into the modulus condition**:
Since \( |z| < 1 \), we need to analyze:
\[ \left| \frac{5\omega - 5}{3 + 5\omega} \right| < 1 \]

8. **Cross-multiply to eliminate the fraction**:
\[ |5\omega - 5| < |3 + 5\omega| \]

9. **Use the triangle inequality**:
\[ |5(\omega - 1)| < |3 + 5\omega| \]

10. **Simplify the inequality**:
\[ 5|\omega - 1| < |3 + 5\omega| \]

### Conclusion:
The derived inequality \( 5|\omega - 1| < |3 + 5\omega| \) gives us a condition that must hold for \( \omega \) based on the initial condition \( |z| < 1 \).