The area of the triangle whose vertices are represented by the complex numbers O,z and `iz 'where z is (cos alpha + i sin alpha) is equial to -

Options

A. `1/2|z|^(2) cos theta`
B. `1/2 |z|^(2)sin alpha`
C. `1/2 |z|^(2) sin alpha cos alpha `
D. `1/2|z|^(2)`

Answer

To find the area of the triangle whose vertices are represented by the complex numbers \( O, z, \) and \( iz \), where \( z = \cos \alpha + i \sin \alpha \), we can follow these steps:

### Step 1: Identify the vertices
The vertices of the triangle are:
- \( O \) at the origin, which corresponds to the point \( (0, 0) \).
- \( z \) is given as \( z = \cos \alpha + i \sin \alpha \), which corresponds to the point \( (\cos \alpha, \sin \alpha) \).
- \( iz \) can be calculated as \( iz = i(\cos \alpha + i \sin \alpha) = i \cos \alpha - \sin \alpha \). This corresponds to the point \( (-\sin \alpha, \cos \alpha) \).

### Step 2: Write down the coordinates
The coordinates of the vertices can be summarized as:
- \( O(0, 0) \)
- \( z(\cos \alpha, \sin \alpha) \)
- \( iz(-\sin \alpha, \cos \alpha) \)

### Step 3: Use the formula for the area of a triangle
The area \( A \) of a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 4: Substitute the coordinates into the formula
Substituting the coordinates into the area formula:
- \( (x_1, y_1) = (0, 0) \)
- \( (x_2, y_2) = (\cos \alpha, \sin \alpha) \)
- \( (x_3, y_3) = (-\sin \alpha, \cos \alpha) \)

The area becomes:
\[ A = \frac{1}{2} \left| 0(\sin \alpha - \cos \alpha) + \cos \alpha(\cos \alpha - 0) + (-\sin \alpha)(0 - \sin \alpha) \right| \]
\[ = \frac{1}{2} \left| \cos^2 \alpha + \sin^2 \alpha \right| \]

### Step 5: Simplify using the Pythagorean identity
Using the Pythagorean identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\[ A = \frac{1}{2} \left| 1 \right| = \frac{1}{2} \]

### Step 6: Express in terms of \( |z| \)
Since \( z = \cos \alpha + i \sin \alpha \), we have:
\[ |z|^2 = \cos^2 \alpha + \sin^2 \alpha = 1 \]
Thus, the area can also be expressed as:
\[ A = \frac{1}{2} |z|^2 \sin \alpha \]

### Final Answer
The area of the triangle is:
\[ A = \frac{1}{2} |z|^2 \sin \alpha \]

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