If `A(z_(1))` and `B(z_(2))` are two points in the Argand plane such that `z_(1)^(2)+z_(2)^(2)+z_(1)z_(2)=0`, then `triangleOAB`, is

Options

A. equilateral
B. isosceles with `angleAOB=pi/2`
C. isosceles with `angleAOB=(2pi)/3`
D. isosceles with `angleAOB=pi/4`

Explanation

To solve the problem, we need to analyze the given condition and determine the nature of triangle OAB formed by points A and B in the Argand plane.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We start with the equation given in the problem:
\[ z_1^2 + z_2^2 + z_1 z_2 = 0 \]

2. **Rearranging the Equation**:
We can rearrange the equation as follows:
\[ z_1^2 + z_1 z_2 + z_2^2 = 0 \]
This can be factored as:
\[ (z_1 + z_2)(z_1 + \omega z_2) = 0 \]
where \(\omega\) is a cube root of unity, specifically \(\omega = e^{i \frac{2\pi}{3}}\).

3. **Finding Relationships Between \(z_1\) and \(z_2\)**:
From the factored form, we can derive:
\[ z_1 = -z_2 \quad \text{or} \quad z_1 = -\omega z_2 \]
This indicates that \(z_1\) can be expressed in terms of \(z_2\) with a rotation.

4. **Expressing \(z_1\) in Polar Form**:
The relationship \(z_1 = -\omega z_2\) can be expressed as:
\[ z_1 = e^{i \frac{2\pi}{3}} z_2 \]
This shows that \(z_1\) is obtained by rotating \(z_2\) by an angle of \(\frac{2\pi}{3}\) radians (or 120 degrees) in the counterclockwise direction.

5. **Analyzing the Triangle OAB**:
- The points O (origin), A (\(z_1\)), and B (\(z_2\)) form triangle OAB.
- The length OA is equal to OB because both points are at the same distance from the origin (since they are related by a rotation).

6. **Determining the Type of Triangle**:
- Since OA = OB and the angle between them (angle AOB) is \(\frac{2\pi}{3}\), triangle OAB is an isosceles triangle.
- The angle AOB is not 90 degrees, confirming that it is not a right triangle.

### Conclusion:
Thus, triangle OAB is an **isosceles triangle**.