Question

The Area of triangle whose vertices are iota, $w$ and $w^2$, where $w$ and $w^2$ are non real cube roots of unity, is :

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{\sqrt{3}}{2}$
• C. 1
• D. $2\sqrt{3}$

Hint:

Remember that the non-real cube roots of unity are defined by:
\[ w, w^2 = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Complex numbers can be represented as coordinates on the Argand plane to calculate geometric areas.
Key Formula or Approach:
The area of a triangle with Cartesian vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:
\[ \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]

Step 2: Detailed Explanation:
Let us identify the three vertices:
- Vertex 1: $z_1 = i \implies (x_1 = 0, y_1 = 1)$
- Vertex 2: $z_2 = w = -\frac{1}{2} + i \frac{\sqrt{3}}{2} \implies \left(x_2 = -\frac{1}{2}, y_2 = \frac{\sqrt{3}}{2}\right)$
- Vertex 3: $z_3 = w^2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \implies \left(x_3 = -\frac{1}{2}, y_3 = -\frac{\sqrt{3}}{2}\right)$
Substitute these coordinate values into the area formula:
\[ \text{Area} = \frac{1}{2} \left| 0 \cdot \left(\frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right)\right) + \left(-\frac{1}{2}\right) \cdot \left(-\frac{\sqrt{3}}{2} - 1\right) + \left(-\frac{1}{2}\right) \cdot \left(1 - \frac{\sqrt{3}}{2}\right) \right| \]
\[ \text{Area} = \frac{1}{2} \left| 0 + \frac{\sqrt{3}}{4} + \frac{1}{2} - \frac{1}{2} + \frac{\sqrt{3}}{4} \right| \]
\[ \text{Area} = \frac{1}{2} \left| \frac{2\sqrt{3}}{4} \right| = \frac{\sqrt{3}}{4} \]
Therefore, the area of the triangle is $\frac{\sqrt{3}}{4}$.

Step 3: Final Answer
The correct option is (A).