Question

The area of the triangle whose vertices are $( i, \omega, \omega^2 )$ is (Where $\omega$ is a complex cube root of unity other than 1, $i$ is an imaginary number)________ sq.units

• A. $\left(\frac{3\sqrt{3}}{4}\right)$
• B. $\left(\frac{\sqrt{3}}{2}\right)$
• C. $\left(\frac{3\sqrt{3}}{2}\right)$
• D. \(\frac{\sqrt{3}}{4}\)

Hint:

$\omega$ and $\omega^2$ always lie on the unit circle with a $120^{\circ}$ angle between them.

Answer 1

The Correct Option is A

Step 1: Identify Coordinates
Vertices in $(\text{Re}, \text{Im})$ form:
$z_1 = i \rightarrow (0, 1)$
$z_2 = \omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \rightarrow \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
$z_3 = \omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \rightarrow \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$.

Step 2: Area Formula
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Step 3: Calculation
\[ \text{Area} = \frac{1}{2} \left| 0 + \left(-\frac{1}{2}\right)\left(-\frac{\sqrt{3}}{2} - 1\right) + \left(-\frac{1}{2}\right)\left(1 - \frac{\sqrt{3}}{2}\right) \right| \] \[ = \frac{1}{2} \left| \frac{\sqrt{3}}{4} + \frac{1}{2} - \frac{1}{2} + \frac{\sqrt{3}}{4} \right| = \frac{1}{2} \left| \frac{2\sqrt{3}}{4} \right| = \frac{\sqrt{3}}{4} \]
Final Answer: (A)