Let `a`, `b`, `c` be distinct complex numbers with `|a|=|b|=|c|=1` and `z_(1)`, `z_(2)` be the roots of the equation `az^(2)+bz+c=0` with `|z_(1)|=1`. Let `P` and `Q` represent the complex numbers `z_(1)` and `z_(2)` in the Argand plane with `/_POQ=theta`, `o^(@) lt 180^(@)` (where `O` being the origin).Then

Options

A. `b^(2)=ac`, `theta=(2pi)/(3)`
B. `theta=(2pi)/(3)`,`PQ=sqrt(3)`
C. `PQ=2sqrt(3)`, `b^(2)=ac`
D. `theta=(pi)/(3)`, `b^(2)=ac`

Explanation

To solve the problem step-by-step, we begin by analyzing the given quadratic equation with roots \( z_1 \) and \( z_2 \).

### Step 1: Understanding the Roots
The roots of the equation \( az^2 + bz + c = 0 \) can be expressed using Vieta's formulas:
- The sum of the roots \( z_1 + z_2 = -\frac{b}{a} \)
- The product of the roots \( z_1 z_2 = \frac{c}{a} \)

### Step 2: Taking Moduli
Given that \( |a| = |b| = |c| = 1 \), we take the moduli of the equations derived from Vieta's formulas:
1. \( |z_1 + z_2| = \left| -\frac{b}{a} \right| = 1 \)
2. \( |z_1 z_2| = \left| \frac{c}{a} \right| = 1 \)

### Step 3: Squaring the Sum of Roots
Next, we square the modulus of the sum of the roots:
\[ |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2 \text{Re}(z_1 \overline{z_2}) = 1 \]
Since \( |z_1| = |z_2| = 1 \), we have:
\[ 1 + 1 + 2 \text{Re}(z_1 \overline{z_2}) = 1 \]
This simplifies to:
\[ 2 + 2 \text{Re}(z_1 \overline{z_2}) = 1 \]
Thus,
\[ \text{Re}(z_1 \overline{z_2}) = -\frac{1}{2} \]

### Step 4: Relating \( z_2 \) to \( z_1 \)
Assuming \( z_2 = z_1 e^{i\theta} \), we substitute:
\[ \text{Re}(z_1 \overline{z_2}) = \text{Re}(z_1 \overline{z_1} e^{-i\theta}) = e^{-i\theta} \]
Thus, we have:
\[ \cos(\theta) = -\frac{1}{2} \]
From this, we find:
\[ \theta = 120^\circ \quad \text{(or } \theta = 240^\circ \text{, but we take the acute angle)} \]

### Step 5: Finding the Length \( PQ \)
Next, we calculate the distance \( PQ = |z_2 - z_1| \):
\[ |z_2 - z_1| = |z_1 e^{i\theta} - z_1| = |z_1| |e^{i\theta} - 1| = |e^{i\theta} - 1| \]
Using \( \theta = 120^\circ \):
\[ |e^{i\theta} - 1| = |e^{i \frac{2\pi}{3}} - 1| = |-\frac{1}{2} + i \frac{\sqrt{3}}{2} - 1| = |-\frac{3}{2} + i \frac{\sqrt{3}}{2}| \]
Calculating the modulus:
\[ \sqrt{\left(-\frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \]

### Final Result
Thus, the angle \( \theta \) is \( 120^\circ \) and the distance \( PQ \) is \( \sqrt{3} \).

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