If `z_1,z_2,z_3` are vertices of a triangle such that `|z_1-z_2|=|z_1-z_3|`, then arg `((2z_1-z_2-z_3)/(z_3-z_2))` is :

Options

A. `pm pi/3`
B. 0
C. `pm pi/2`
D. `pm pi/6`

Answer

To solve the problem, we need to find the argument of the complex expression \(\frac{2z_1 - z_2 - z_3}{z_3 - z_2}\) given that \(|z_1 - z_2| = |z_1 - z_3|\).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
The condition \(|z_1 - z_2| = |z_1 - z_3|\) implies that point \(z_1\) is equidistant from points \(z_2\) and \(z_3\). This means that \(z_1\) lies on the perpendicular bisector of the segment joining \(z_2\) and \(z_3\).

2. **Rewriting the Expression**:
We need to simplify the expression \(\frac{2z_1 - z_2 - z_3}{z_3 - z_2}\). We can rewrite the numerator:
\[ 2z_1 - z_2 - z_3 = (z_1 - z_2) + (z_1 - z_3) \]

3. **Expressing the Argument**:
The argument of a quotient of complex numbers can be expressed as the difference of their arguments:
\[ \arg\left(\frac{2z_1 - z_2 - z_3}{z_3 - z_2}\right) = \arg(2z_1 - z_2 - z_3) - \arg(z_3 - z_2) \]

4. **Finding Each Argument**:
- Let \(A = z_1 - z_2\) and \(B = z_1 - z_3\). Then:
\[ \arg(2z_1 - z_2 - z_3) = \arg(A + B) \]
- Since \(|A| = |B|\) (from the given condition), we can say that \(A\) and \(B\) are of equal magnitude but may differ in direction.

5. **Using the Geometry of the Situation**:
Since \(z_1\) lies on the perpendicular bisector of \(z_2\) and \(z_3\), the angle between \(A\) and \(B\) is \(180^\circ\) (or \(\pi\) radians). Thus, we can express:
\[ \arg(A + B) = \arg(A) + \arg(B) + \pi \]

6. **Calculating the Final Argument**:
Thus, we have:
\[ \arg\left(\frac{2z_1 - z_2 - z_3}{z_3 - z_2}\right) = \arg(A + B) - \arg(z_3 - z_2) \]
Given that \(z_3 - z_2\) is simply the vector from \(z_2\) to \(z_3\), we can conclude that:
\[ \arg\left(\frac{2z_1 - z_2 - z_3}{z_3 - z_2}\right) = \frac{\pi}{2} \text{ or } -\frac{\pi}{2} \]
depending on the orientation of the triangle.

### Final Answer:
The argument \(\arg\left(\frac{2z_1 - z_2 - z_3}{z_3 - z_2}\right)\) is \(\pm \frac{\pi}{2}\).