If `lambda in R` such that the origin and the non-real roots of the equation `2z^(2)+2z+lambda=0` form the vertices of an equilateral triangle in the argand plane, then `(1)/(lambda)` is equal to

Answer

To solve the problem, we need to find the value of \( \frac{1}{\lambda} \) given that the origin and the non-real roots of the equation \( 2z^2 + 2z + \lambda = 0 \) form the vertices of an equilateral triangle in the Argand plane.

### Step-by-step Solution:

1. **Identify the roots of the equation:**
The given quadratic equation is \( 2z^2 + 2z + \lambda = 0 \). We can use the quadratic formula to find the roots:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 \), \( b = 2 \), and \( c = \lambda \). Plugging in these values, we get:
\[ z = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot \lambda}}{2 \cdot 2} = \frac{-2 \pm \sqrt{4 - 8\lambda}}{4} \]

2. **Condition for non-real roots:**
For the roots to be non-real, the discriminant must be negative:
\[ 4 - 8\lambda < 0 \implies 8\lambda > 4 \implies \lambda > \frac{1}{2} \]

3. **Sum and product of the roots:**
The sum of the roots \( z_1 + z_2 \) is given by:
\[ z_1 + z_2 = -\frac{b}{a} = -\frac{2}{2} = -1 \]
The product of the roots \( z_1 z_2 \) is given by:
\[ z_1 z_2 = \frac{c}{a} = \frac{\lambda}{2} \]

4. **Condition for equilateral triangle:**
The condition for the points \( O \) (origin), \( z_1 \), and \( z_2 \) to form an equilateral triangle in the Argand plane is:
\[ z_1^2 + z_2^2 = z_1 z_2 \]
We can express \( z_1^2 + z_2^2 \) using the identity:
\[ z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1 z_2 \]
Substituting the values we have:
\[ z_1^2 + z_2^2 = (-1)^2 - 2\left(\frac{\lambda}{2}\right) = 1 - \lambda \]

5. **Setting up the equation:**
Now we set up the equation from the equilateral triangle condition:
\[ 1 - \lambda = \frac{\lambda}{2} \]

6. **Solving for \( \lambda \):**
Rearranging the equation gives:
\[ 1 = \lambda + \frac{\lambda}{2} \implies 1 = \frac{2\lambda + \lambda}{2} \implies 1 = \frac{3\lambda}{2} \]
Multiplying both sides by 2:
\[ 2 = 3\lambda \implies \lambda = \frac{2}{3} \]

7. **Finding \( \frac{1}{\lambda} \):**
Finally, we calculate \( \frac{1}{\lambda} \):
\[ \frac{1}{\lambda} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \]

### Final Answer:
\[ \frac{1}{\lambda} = \frac{3}{2} \]