Question

If \( |z_1|=|z_2|=|z_3|=1 \) and \( z_1+z_2+z_3=0 \), then the area of the triangle whose vertices are \( z_1,z_2,z_3 \) is:

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

Hint:

If unit complex numbers sum to zero: \begin{itemize} \item They form cube roots of unity. \item Triangle is equilateral. \end{itemize}

Answer 1

The Correct Option is A

Concept: Three unit complex numbers summing to zero lie at vertices of an equilateral triangle on unit circle. Step 1: {\color{red}Interpret geometrically.} Points are cube roots of unity: \[ 1, \omega, \omega^2 \] Form equilateral triangle. Step 2: {\color{red}Find side length.} Distance between roots: \[ |1 - \omega| = \sqrt{3} \] So side = \( \sqrt{3} \). Step 3: {\color{red}Area formula.} \[ \text{Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \cdot 3 = \frac{3\sqrt{3}}{4} \]