The area of the triangle (in square units) whose vertices are `i`, `omega` and `omega^(2)` where `i=sqrt(-1)` and `omega, omega^(2)` are complex cube roots of unity, is

Options

A. `(3sqrt(3))/2`
B. `(3sqrt(3))/(4)`
C. 0
D. `sqrt(3)/4`

Answer

To find the area of the triangle with vertices at \(i\), \(\omega\), and \(\omega^2\) where \(i = \sqrt{-1}\) and \(\omega\) and \(\omega^2\) are the complex cube roots of unity, we can follow these steps:

### Step 1: Identify the vertices
The vertices of the triangle are given as:
- \(z_1 = i = 0 + 1i\)
- \(z_2 = \omega = \frac{1}{2} - \frac{\sqrt{3}}{2}i\)
- \(z_3 = \omega^2 = \frac{1}{2} + \frac{\sqrt{3}}{2}i\)

### Step 2: Convert complex numbers to Cartesian coordinates
We can express the vertices in Cartesian coordinates:
- \(A(0, 1)\) for \(z_1\)
- \(B\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)\) for \(z_2\)
- \(C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) for \(z_3\)

### Step 3: Use the formula for the area of a triangle
The area \(A\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using 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:
- \(x_1 = 0\), \(y_1 = 1\)
- \(x_2 = \frac{1}{2}\), \(y_2 = -\frac{\sqrt{3}}{2}\)
- \(x_3 = \frac{1}{2}\), \(y_3 = \frac{\sqrt{3}}{2}\)

The area becomes:
\[ A = \frac{1}{2} \left| 0\left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) + \frac{1}{2}\left(\frac{\sqrt{3}}{2} - 1\right) + \frac{1}{2}\left(1 - \left(-\frac{\sqrt{3}}{2}\right)\right) \right| \]

### Step 5: Simplify the expression
Calculating each term:
- The first term is \(0\).
- The second term: \(\frac{1}{2}\left(\frac{\sqrt{3}}{2} - 1\right) = \frac{\sqrt{3}}{4} - \frac{1}{2}\).
- The third term: \(\frac{1}{2}\left(1 + \frac{\sqrt{3}}{2}\right) = \frac{1}{2} + \frac{\sqrt{3}}{4}\).

Now combine these:
\[ A = \frac{1}{2} \left| \left(\frac{\sqrt{3}}{4} - \frac{1}{2}\right) + \left(\frac{1}{2} + \frac{\sqrt{3}}{4}\right) \right| \]
\[ = \frac{1}{2} \left| \frac{\sqrt{3}}{4} - \frac{1}{2} + \frac{1}{2} + \frac{\sqrt{3}}{4} \right| \]
\[ = \frac{1}{2} \left| \frac{2\sqrt{3}}{4} \right| = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \]

### Step 6: Final area calculation
Thus, the area of the triangle is:
\[ A = \frac{\sqrt{3}}{4} \text{ square units} \]

### Conclusion
The area of the triangle whose vertices are \(i\), \(\omega\), and \(\omega^2\) is \(\frac{\sqrt{3}}{4}\) square units.