If the complex number `A(z_(1)),B(z_(2))` and origin forms an isosceles triangle such that ` angle(AOB) = (2pi)/3 `,then `(z_(1)^(2)+z_(2)^(2) +4z_(1)z_(2))/(z_(1)z_(2))` equals ________

Answer

To solve the problem, we need to find the value of the expression

\[ \frac{z_1^2 + z_2^2 + 4z_1z_2}{z_1z_2} \]

given that the complex numbers \( A(z_1) \), \( B(z_2) \), and the origin form an isosceles triangle with \( \angle AOB = \frac{2\pi}{3} \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
Since \( AOB \) is an isosceles triangle and \( \angle AOB = \frac{2\pi}{3} \), we know that \( |z_1| = |z_2| \). Let \( |z_1| = |z_2| = r \).

2. **Using the Angle**:
We can express \( z_1 \) and \( z_2 \) in polar form:
\[ z_1 = r e^{i\theta}, \quad z_2 = r e^{i(\theta + \frac{2\pi}{3})} \]

3. **Calculating \( z_1^2 \) and \( z_2^2 \)**:
\[ z_1^2 = r^2 e^{2i\theta}, \quad z_2^2 = r^2 e^{2i(\theta + \frac{2\pi}{3})} \]

4. **Finding \( z_1^2 + z_2^2 \)**:
\[ z_2^2 = r^2 e^{2i\theta} e^{\frac{4\pi i}{3}} = r^2 e^{2i\theta} \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \]
Therefore,
\[ z_1^2 + z_2^2 = r^2 e^{2i\theta} + r^2 e^{2i\theta} \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \]
Simplifying this gives:
\[ z_1^2 + z_2^2 = r^2 e^{2i\theta} \left(1 - \frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = r^2 e^{2i\theta} \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \]

5. **Calculating \( z_1z_2 \)**:
\[ z_1 z_2 = r^2 e^{i\theta} e^{i(\theta + \frac{2\pi}{3})} = r^2 e^{2i\theta} e^{\frac{2\pi i}{3}} \]

6. **Substituting into the Expression**:
Now we substitute \( z_1^2 + z_2^2 \) and \( z_1 z_2 \) into the expression:
\[ \frac{z_1^2 + z_2^2 + 4z_1z_2}{z_1z_2} = \frac{z_1^2 + z_2^2}{z_1z_2} + 4 \]

7. **Finding \( \frac{z_1^2 + z_2^2}{z_1z_2} \)**:
Since \( z_1^2 + z_2^2 = -z_1z_2 \) (from the isosceles triangle property),
\[ \frac{z_1^2 + z_2^2}{z_1z_2} = -1 \]

8. **Final Calculation**:
Therefore,
\[ \frac{z_1^2 + z_2^2 + 4z_1z_2}{z_1z_2} = -1 + 4 = 3 \]

### Conclusion:
The value of the expression is

\[ \boxed{3} \]