The triangle formed by complex numbers `z ,i z ,i^2z` is
(a)Equilateral
(b) Isosceles
(c)Right angle
(d) Scalene

Options

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

Answer

To determine the type of triangle formed by the complex numbers \( z, iz, i^2z \), we will follow these steps:

### Step 1: Define the complex number \( z \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Find the coordinates of the points
- The point corresponding to \( z \) is \( A(x, y) \).
- The point corresponding to \( iz \) is \( B(iy) = (-y, x) \).
- The point corresponding to \( i^2z \) is \( C(i^2z) = -z = -x - iy \), which corresponds to \( C(-x, -y) \).

### Step 3: Calculate the distances between the points
1. **Distance \( AB \)**:
\[ AB = \sqrt{((-y) - x)^2 + (x - y)^2} = \sqrt{(-y - x)^2 + (x - y)^2} \]
Simplifying:
\[ AB = \sqrt{(y + x)^2 + (x - y)^2} \]

2. **Distance \( BC \)**:
\[ BC = \sqrt{((-x) - (-y))^2 + ((-y) - x)^2} = \sqrt{(-x + y)^2 + (-y - x)^2} \]
Simplifying:
\[ BC = \sqrt{(y - x)^2 + (-y - x)^2} \]

3. **Distance \( CA \)**:
\[ CA = \sqrt{(x - (-x))^2 + (y - (-y))^2} = \sqrt{(2x)^2 + (2y)^2} = \sqrt{4(x^2 + y^2)} = 2\sqrt{x^2 + y^2} \]

### Step 4: Analyze the triangle
Now we have the distances:
- \( AB \)
- \( BC \)
- \( CA = 2\sqrt{x^2 + y^2} \)

To check if the triangle is isosceles, we can see if \( AB = BC \).

### Step 5: Check for right angle
To check if the triangle is a right triangle, we can use the Pythagorean theorem:
\[ AB^2 + BC^2 = CA^2 \]

### Conclusion
From the calculations, we find that:
- \( AB = BC \) indicates that the triangle is isosceles.
- The condition \( AB^2 + BC^2 = CA^2 \) indicates that the triangle is also a right triangle.

Thus, the triangle formed by the complex numbers \( z, iz, i^2z \) is both isosceles and right-angled.

### Final Answer
The correct options are (b) Isosceles and (c) Right angle.

---