Let the complex numbers `Z_(1), Z_(2) and Z_(3)` are the vertices A, B and C respectively of an isosceles right - angled triangle ABC with right angle at C, then the value of `((Z_(1)-Z_(2))^(2))/((Z_(1)-Z_(3))(Z_(3)-Z_(2)))` is equal to

Options

A. 2
B. 1
C. `-2`
D. `-1`

Answer

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

\[ \frac{(Z_1 - Z_2)^2}{(Z_1 - Z_3)(Z_3 - Z_2)} \]

where \(Z_1\), \(Z_2\), and \(Z_3\) are the vertices of an isosceles right triangle with the right angle at vertex \(C\).

### Step 1: Understand the Geometry
Since triangle \(ABC\) is isosceles and right-angled at \(C\), we can denote the vertices in the complex plane as follows:
- Let \(Z_1 = a + bi\) (vertex \(A\))
- Let \(Z_2 = a - bi\) (vertex \(B\))
- Let \(Z_3 = a + 0i\) (vertex \(C\))

Here, \(b\) is the length of the legs of the triangle, and \(a\) is the x-coordinate of the right angle vertex \(C\).

### Step 2: Calculate \(Z_1 - Z_2\)
\[ Z_1 - Z_2 = (a + bi) - (a - bi) = 2bi \]

### Step 3: Calculate \(Z_1 - Z_3\) and \(Z_3 - Z_2\)
\[ Z_1 - Z_3 = (a + bi) - (a + 0i) = bi \]
\[ Z_3 - Z_2 = (a + 0i) - (a - bi) = bi \]

### Step 4: Substitute into the Expression
Now we substitute these values into our expression:
\[ \frac{(Z_1 - Z_2)^2}{(Z_1 - Z_3)(Z_3 - Z_2)} = \frac{(2bi)^2}{(bi)(bi)} \]

### Step 5: Simplify the Expression
Calculating the numerator:
\[ (2bi)^2 = 4b^2(-1) = -4b^2 \]

Calculating the denominator:
\[ (bi)(bi) = b^2(-1) = -b^2 \]

Now substituting back:
\[ \frac{-4b^2}{-b^2} = 4 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{4} \]