Let ` z_(1) , z_(2)` be two non - zero complex numbers such that ` z_(1)^(2) - 2 z _(1) z_(2) + 2 z_(2)^(2)` = 0 then the triangle OAB, where O is origin and A, B are ` z_(1) and z_(2)` is

Options

A. isosceles right angled
B. equilateral
C. isosceles but not right angled
D. only right angled

Answer

To solve the problem, we start with the given equation involving the complex numbers \( z_1 \) and \( z_2 \):

\[ z_1^2 - 2z_1 z_2 + 2z_2^2 = 0 \]

### Step 1: Rearranging the Equation
We can rearrange the equation to isolate the terms involving \( z_1 \):

\[ z_1^2 - 2z_1 z_2 + 2z_2^2 = 0 \]

This can be treated as a quadratic equation in \( z_1 \).

### Step 2: Using the Quadratic Formula
The quadratic formula states that for an equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our case, \( a = 1 \), \( b = -2z_2 \), and \( c = 2z_2^2 \). Plugging these values into the quadratic formula gives:

\[ z_1 = \frac{2z_2 \pm \sqrt{(-2z_2)^2 - 4 \cdot 1 \cdot 2z_2^2}}{2 \cdot 1} \]

### Step 3: Simplifying the Discriminant
Now we simplify the discriminant:

\[ (-2z_2)^2 - 4 \cdot 1 \cdot 2z_2^2 = 4z_2^2 - 8z_2^2 = -4z_2^2 \]

### Step 4: Finding the Roots
Substituting back into the quadratic formula, we have:

\[ z_1 = \frac{2z_2 \pm \sqrt{-4z_2^2}}{2} \]

This simplifies to:

\[ z_1 = \frac{2z_2 \pm 2i z_2}{2} = z_2(1 \pm i) \]

### Step 5: Relation Between \( z_1 \) and \( z_2 \)
Thus, we find that:

\[ z_1 = z_2(1 + i) \quad \text{or} \quad z_1 = z_2(1 - i) \]

### Step 6: Analyzing the Triangle OAB
Now, we need to analyze the triangle formed by the points \( O \) (origin), \( A \) (point represented by \( z_1 \)), and \( B \) (point represented by \( z_2 \)).

1. **Coordinates of Points**:
- \( O(0, 0) \)
- \( A(z_2(1 + i)) = (z_2, z_2) \)
- \( B(z_2) = (z_2, 0) \)

2. **Calculating Distances**:
- The distance \( OA = |z_1| = |z_2| \sqrt{2} \) (since \( |1 + i| = \sqrt{2} \))
- The distance \( OB = |z_2| \)
- The distance \( AB = |z_1 - z_2| = |z_2(1 + i) - z_2| = |z_2 i| = |z_2| \)

### Step 7: Conclusion
Since \( OA = |z_2| \sqrt{2} \) and \( OB = |z_2| \), we see that \( OA \) is longer than \( OB \), and the triangle formed is a right-angled triangle with \( O \) as the right angle.

### Final Answer
Thus, the triangle OAB is a **right-angled triangle**.

---