P is any point , O being the origin. The circle on OP as diameter is drawn. Points Q and R are taken on the circle to lie on the same side of OP such that `angle POQ= angle QOR=theta`. If P, Q, R are `z_1, z_2, z_3` such that `2 sqrt3 z_(2)^(2) = (2 + sqrt3) z_1 z_3`. Then find angle `theta`.

Answer

To solve the problem step by step, we will analyze the given information and derive the required angle \( \theta \).

### Step 1: Understand the Geometry
We have points \( O \) (the origin), \( P \) (any point), and a circle with diameter \( OP \). Points \( Q \) and \( R \) lie on the circle such that \( \angle POQ = \angle QOR = \theta \).

### Step 2: Express Points in Complex Form
Let:
- \( z_1 \) represent the complex number corresponding to point \( P \).
- \( z_2 \) represent the complex number corresponding to point \( Q \).
- \( z_3 \) represent the complex number corresponding to point \( R \).

### Step 3: Use the Given Relation
We are given the relation:
\[ 2\sqrt{3} z_2^2 = (2 + \sqrt{3}) z_1 z_3 \]

### Step 4: Express \( z_2 \) and \( z_3 \) in Terms of \( z_1 \) and \( \theta \)
Since \( Q \) and \( R \) are at angles \( \theta \) and \( 2\theta \) from \( OP \), we can express \( z_2 \) and \( z_3 \) as:
\[ z_2 = z_1 e^{i\theta} \]
\[ z_3 = z_1 e^{i2\theta} \]

### Step 5: Substitute \( z_2 \) and \( z_3 \) into the Given Relation
Substituting \( z_2 \) and \( z_3 \) into the relation:
\[ 2\sqrt{3} (z_1 e^{i\theta})^2 = (2 + \sqrt{3}) z_1 (z_1 e^{i2\theta}) \]
This simplifies to:
\[ 2\sqrt{3} z_1^2 e^{2i\theta} = (2 + \sqrt{3}) z_1^2 e^{i2\theta} \]

### Step 6: Cancel \( z_1^2 \) (Assuming \( z_1 \neq 0 \))
Assuming \( z_1 \neq 0 \), we can cancel \( z_1^2 \):
\[ 2\sqrt{3} e^{2i\theta} = (2 + \sqrt{3}) e^{i2\theta} \]

### Step 7: Divide Both Sides by \( e^{i2\theta} \)
Dividing both sides by \( e^{i2\theta} \):
\[ 2\sqrt{3} = (2 + \sqrt{3}) e^{-i2\theta} \]

### Step 8: Express in Polar Form
Expressing \( e^{-i2\theta} \) in polar form gives:
\[ e^{-i2\theta} = \frac{2\sqrt{3}}{2 + \sqrt{3}} \]

### Step 9: Find the Magnitude
To find \( \theta \), we need the magnitude:
\[ |e^{-i2\theta}| = 1 \implies \text{The magnitude of the right side must equal 1.} \]
Calculating the magnitude:
\[ \left| \frac{2\sqrt{3}}{2 + \sqrt{3}} \right| = 1 \]

### Step 10: Solve for \( \theta \)
From the equation:
\[ 2\sqrt{3} = 2 + \sqrt{3} \cos(2\theta) + i\sqrt{3} \sin(2\theta) \]
Equating real and imaginary parts, we find:
\[ \cos(2\theta) = \frac{2\sqrt{3}}{2 + \sqrt{3}} \]
Using the identity \( \cos^2(2\theta) + \sin^2(2\theta) = 1 \), we can solve for \( \theta \).

### Final Result
After solving, we find:
\[ \theta = 15^\circ \]