Complex Numbers
Modulus of product from area condition
MJAT_TS1_P1
Grade 12
Question:
Let $z_1, z_2, z_3$ be three distinct complex numbers such that $|z_1| = |z_2| = |z_3| = 4$ and
$$z_1(z_2 - z_3) - \bar{z}_1(\bar{z}_2 - \bar{z}_3) + z_2\bar{z}_3 - \bar{z}_3 z_2 = 48$$
Then $|(z_1 - z_2)(z_2 - z_3)(z_3 - z_1)|$ is equal to:
Step-by-Step Solution
Key Concept: The given expression is $2i \cdot \text{Im}(z_1(\bar{z}_2 - \bar{z}_3)) + 2i\,\text{Im}(z_2\bar{z}_3)$. Rewrite using the determinant form for the area of triangle $z_1z_2z_3$: $|\text{Area}| = \frac{1}{4}\left|\begin{vmatrix}z_1 & \bar{z}_1 & 1\\ z_2 & \bar{z}_2 & 1 \\ z_3 & \bar{z}_3 & 1\end{vmatrix}\right|$. Relate this determinant to 48.
From the condition, $2\cdot\text{Im}$ of determinant $= 48$, so $\text{Area}(\triangle z_1z_2z_3) = 6$. By the formula $\text{Area} = \frac{abc}{4R}$: $6 = \frac{abc}{4\times 4}$, so $abc = |(z_1-z_2)(z_2-z_3)(z_3-z_1)| = 6\times 32 = 192$.
Correct Answer: 192