Let `z_(1)`, `z_(2)`, `z_(3)` are the vertices of `DeltaABC`, respectively, such that `(z_(3)-z_(2))/(z_(1)-z_(2))` is purely imaginery number. A square on side `AC` is drawn outwardly. `P(z_(4))` is the centre of square, then

Options

A. `|z_(1)-z_(2)|=|z_(2)-z_(4)|`
B. `arg((z_(1)-z_(2))/(z_(4)-z_(2)))+arg((z_(3)-z_(2))/(z_(4)-z_(2)))=+(pi)/(2)`
C. `arg((z_(1)-z_(2))/(z_(4)-z_(2)))+arg((z_(3)-z_(2))/(z_(4)-z_(2)))=0`
D. `z_(1)`, `z_(2)`, `z_(3)` and `z_(4)` lie on a circle

Explanation

To solve the problem, we need to analyze the given conditions and apply properties of complex numbers and geometry. Let's break it down step by step.

### Step 1: Understanding the Condition
We are given that:

\[ \frac{z_3 - z_2}{z_1 - z_2} \text{ is purely imaginary.} \]

This means that the argument (angle) of the complex number \(\frac{z_3 - z_2}{z_1 - z_2}\) is \(\frac{\pi}{2}\) or \(-\frac{\pi}{2}\). Therefore, we can write:

\[ \text{arg}(z_3 - z_2) - \text{arg}(z_1 - z_2) = \frac{\pi}{2} \quad \text{(or -}\frac{\pi}{2}\text{)}. \]

### Step 2: Establishing the Geometry
Since a square is drawn outwardly on side \(AC\), we denote the vertices of the square as \(A\), \(C\), and \(P\) (the center of the square). The center \(P\) can be expressed in terms of \(A\) and \(C\).

### Step 3: Finding the Center of the Square
The center \(P\) of the square on side \(AC\) can be calculated as:

\[ P = \frac{A + C}{2} + \frac{i}{2} (C - A), \]

where \(i\) represents the imaginary unit, indicating a rotation by \(90^\circ\).

### Step 4: Using the Properties of Cyclic Quadrilaterals
Since \(A\), \(B\), \(C\), and \(P\) form a cyclic quadrilateral, we can use the property that the opposite angles sum to \(180^\circ\). This gives us:

\[ \angle ABP + \angle ACP = 180^\circ. \]

### Step 5: Finding the Argument Relations
From the cyclic nature, we can express the arguments:

1. Let \(\text{arg}(z_1 - z_2) = -\alpha\).
2. Let \(\text{arg}(z_3 - z_2) = \alpha\).

Now, we can write:

\[ \text{arg}(z_1 - z_2) + \text{arg}(z_3 - z_2) = 0. \]

### Step 6: Finalizing the Argument
By adding the arguments, we have:

\[ \text{arg}(z_1 - z_2) + \text{arg}(z_3 - z_2) = -\alpha + \alpha = 0. \]

Thus, we conclude that:

\[ \text{arg}(z_1 - z_2) + \text{arg}(z_3 - z_2) = 0. \]

### Conclusion
The final result is:

\[ \text{arg}(z_1 - z_2) + \text{arg}(z_3 - z_2) = 0. \]