One vertex of an equilateral triangle is at the origin and the other two vertices are given by ` 2 z^(2) + 2 z + k = 0 ` then k is

Options

A. `2//3`
B. 1
C. 2
D. `-1`

Answer

To solve the problem, we need to find the value of \( k \) such that the vertices of the equilateral triangle, with one vertex at the origin, satisfy the equation \( 2z^2 + 2z + k = 0 \).

### Step-by-Step Solution:

1. **Identify the Vertices**:
Let the vertices of the equilateral triangle be \( 0 \) (the origin), \( z_1 \), and \( z_2 \). The equation given is \( 2z^2 + 2z + k = 0 \), which represents the other two vertices \( z_1 \) and \( z_2 \).

2. **Use the Properties of Equilateral Triangles**:
In an equilateral triangle, the angle between any two vertices (from the origin) is \( \frac{\pi}{3} \) radians. Therefore, we can express the relationship between \( z_1 \) and \( z_2 \) as:
\[ z_1 = z_2 e^{i \frac{\pi}{3}} = z_2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \]

3. **Sum of Roots**:
According to Vieta's formulas, the sum of the roots \( z_1 + z_2 \) is given by:
\[ z_1 + z_2 = -\frac{b}{a} = -\frac{2}{2} = -1 \]
Substituting \( z_1 \):
\[ z_2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) + z_2 = -1 \]
Simplifying this:
\[ z_2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} + 1 \right) = -1 \]
\[ z_2 \left( \frac{3}{2} + i \frac{\sqrt{3}}{2} \right) = -1 \]
Thus,
\[ z_2 = -\frac{2}{3 + i \sqrt{3}} \]

4. **Product of Roots**:
The product of the roots \( z_1 z_2 \) is given by:
\[ z_1 z_2 = \frac{c}{a} = \frac{k}{2} \]
Substituting \( z_1 \):
\[ z_1 z_2 = \left( z_2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \right) z_2 = z_2^2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \]
Therefore,
\[ z_2^2 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{k}{2} \]

5. **Calculate \( z_2^2 \)**:
First, calculate \( z_2^2 \):
\[ z_2 = -\frac{2}{3 + i \sqrt{3}} \]
To simplify \( z_2 \), multiply the numerator and denominator by the conjugate:
\[ z_2 = -\frac{2(3 - i \sqrt{3})}{(3 + i \sqrt{3})(3 - i \sqrt{3})} = -\frac{2(3 - i \sqrt{3})}{9 + 3} = -\frac{2(3 - i \sqrt{3})}{12} = -\frac{1}{6}(3 - i \sqrt{3}) \]
Now, square \( z_2 \):
\[ z_2^2 = \left(-\frac{1}{6}(3 - i \sqrt{3})\right)^2 = \frac{1}{36}(9 - 6i\sqrt{3} - 3) = \frac{1}{36}(6 - 6i\sqrt{3}) = \frac{1}{6}(1 - i\sqrt{3}) \]

6. **Substitute Back to Find \( k \)**:
Substitute \( z_2^2 \) back into the product equation:
\[ \frac{1}{6}(1 - i\sqrt{3}) \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{k}{2} \]
Simplifying the left-hand side:
\[ \frac{1}{6} \left( \frac{1}{2} - \frac{\sqrt{3}}{2} i + \frac{\sqrt{3}}{2} i - \frac{3}{2} \right) = \frac{1}{6} \left( -1 \right) = -\frac{1}{6} \]
Thus:
\[ -\frac{1}{6} = \frac{k}{2} \implies k = -\frac{1}{3} \]

### Final Answer:
The value of \( k \) is \( -\frac{1}{3} \).