If `z_1 and z_2` are the roots of the equation `z^2+az+12=0 and z_1 , z_2` forms an equilateral triangle with origin. Find `absa`

Answer

To solve the problem, we need to find the value of \(|a|\) given that \(z_1\) and \(z_2\) are the roots of the quadratic equation \(z^2 + az + 12 = 0\) and that \(z_1\) and \(z_2\) form an equilateral triangle with the origin.

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the equation \(z^2 + az + 12 = 0\) can be expressed using the quadratic formula:
\[ z_{1,2} = \frac{-a \pm \sqrt{a^2 - 4 \cdot 12}}{2} \]
This simplifies to:
\[ z_{1,2} = \frac{-a \pm \sqrt{a^2 - 48}}{2} \]

2. **Geometric Interpretation**:
Since \(z_1\) and \(z_2\) form an equilateral triangle with the origin, the distance from the origin to both \(z_1\) and \(z_2\) must be equal. This means:
\[ |z_1| = |z_2| \]

3. **Expressing the Roots in Polar Form**:
Let’s express \(z_1\) and \(z_2\) in polar form:
\[ z_1 = r e^{i\theta}, \quad z_2 = r e^{i(\theta + \frac{2\pi}{3})} \]
Here, \(r = |z_1| = |z_2|\) and the angle difference of \(\frac{2\pi}{3}\) corresponds to the angles of an equilateral triangle.

4. **Calculating the Product of the Roots**:
From Vieta's formulas, we know that:
\[ z_1 z_2 = 12 \]
Substituting the polar forms:
\[ (r e^{i\theta})(r e^{i(\theta + \frac{2\pi}{3})}) = r^2 e^{i(2\theta + \frac{2\pi}{3})} = 12 \]
This gives us:
\[ r^2 = 12 \quad \text{and} \quad 2\theta + \frac{2\pi}{3} = 0 \quad (\text{mod } 2\pi) \]

5. **Solving for \(r\)**:
From \(r^2 = 12\), we find:
\[ r = \sqrt{12} = 2\sqrt{3} \]

6. **Finding \(\theta\)**:
From \(2\theta + \frac{2\pi}{3} = 0\):
\[ 2\theta = -\frac{2\pi}{3} \implies \theta = -\frac{\pi}{3} \]

7. **Finding the Roots**:
Now substituting back:
\[ z_1 = 2\sqrt{3} e^{-i\frac{\pi}{3}}, \quad z_2 = 2\sqrt{3} e^{i\frac{\pi}{3}} \]

8. **Calculating \(z_1 + z_2\)**:
\[ z_1 + z_2 = 2\sqrt{3} \left(e^{-i\frac{\pi}{3}} + e^{i\frac{\pi}{3}}\right) = 2\sqrt{3} \cdot 2 \cdot \cos\left(\frac{\pi}{3}\right) = 2\sqrt{3} \cdot 2 \cdot \frac{1}{2} = 2\sqrt{3} \]

9. **Using Vieta's Formulas**:
From Vieta's formulas, we know:
\[ z_1 + z_2 = -a \implies -a = 2\sqrt{3} \implies a = -2\sqrt{3} \]

10. **Finding \(|a|\)**:
Thus, the absolute value of \(a\) is:
\[ |a| = | -2\sqrt{3} | = 2\sqrt{3} \]

### Final Answer:
\[ |a| = 2\sqrt{3} \]