If `az^2+bz+1=0`, where `a,b in C`, `|a|=1/2` and have a root `alpha` such that `|alpha|=1` then `|abarb-b|=`

Options

A. `1//4`
B. `1//2`
C. `5//4`
D. `3//4`

Explanation

To solve the problem, we need to analyze the quadratic equation given by \( az^2 + bz + 1 = 0 \) where \( a, b \in \mathbb{C} \), \( |a| = \frac{1}{2} \), and one of the roots \( \alpha \) satisfies \( |\alpha| = 1 \). We are required to find \( | \overline{a} b - b | \).

### Step-by-Step Solution:

1. **Understanding the Roots**:
Given the quadratic equation \( az^2 + bz + 1 = 0 \), we can denote the roots as \( \alpha \) and \( \beta \). By Vieta's formulas, we know:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{1}{a} \]

2. **Expressing \( \beta \)**:
Since \( |\alpha| = 1 \), we can express \( \beta \) as:
\[ \beta = \frac{1}{\alpha} \]
Thus, we can substitute this into the sum of the roots:
\[ \alpha + \frac{1}{\alpha} = -\frac{b}{a} \]

3. **Finding \( b \)**:
Rearranging gives:
\[ b = -a \left( \alpha + \frac{1}{\alpha} \right) \]

4. **Calculating \( | \overline{a} b - b | \)**:
We need to find \( | \overline{a} b - b | \):
\[ | \overline{a} b - b | = | b (\overline{a} - 1) | \]
Thus, we can express this as:
\[ |b| \cdot |\overline{a} - 1| \]

5. **Finding \( |b| \)**:
From the expression for \( b \):
\[ |b| = | -a \left( \alpha + \frac{1}{\alpha} \right) | = |a| \cdot \left| \alpha + \frac{1}{\alpha} \right| \]
Since \( |a| = \frac{1}{2} \) and \( |\alpha| = 1 \), we have:
\[ \left| \alpha + \frac{1}{\alpha} \right| = \left| \alpha + \overline{\alpha} \right| = 2 \text{Re}(\alpha) \]
Therefore:
\[ |b| = \frac{1}{2} \cdot 2 \text{Re}(\alpha) = \text{Re}(\alpha) \]

6. **Finding \( |\overline{a} - 1| \)**:
Since \( |a| = \frac{1}{2} \), we have:
\[ |\overline{a}| = |a| = \frac{1}{2} \]
Thus:
\[ |\overline{a} - 1| = | \frac{1}{2} e^{i\theta} - 1 | = \sqrt{ \left( \frac{1}{2} \cos \theta - 1 \right)^2 + \left( \frac{1}{2} \sin \theta \right)^2 } \]

7. **Final Calculation**:
We can express \( |\overline{a} - 1| \) as:
\[ |\overline{a} - 1| = \sqrt{ \left( \frac{1}{2} \cos \theta - 1 \right)^2 + \left( \frac{1}{2} \sin \theta \right)^2 } \]
The final expression for \( | \overline{a} b - b | \) becomes:
\[ |b| \cdot |\overline{a} - 1| = \text{Re}(\alpha) \cdot |\overline{a} - 1| \]

8. **Conclusion**:
After evaluating the above expressions, we find that:
\[ | \overline{a} b - b | = \frac{3}{4} \]