Complex Numbers
Circulant determinant and equilateral triangle
MJAT_TS5_P1
Grade 12
Question:
Let $z_1,z_2,z_3$ be nonzero complex numbers with $|z_1|=|z_2|=|z_3|$, $z_2\neq 1$, $a=|z_1|$, $b=|z_2|$, $c=|z_3|$. Let $\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=0$. Then:
A) $\arg\dfrac{z_3}{z_2}=\arg\dfrac{z_3-z_1}{z_2-z_1}$
B) Orthocentre of triangle formed by $z_1,z_2,z_3$ is $z_1+z_2+z_3$
C) If triangle formed by $z_1,z_2,z_3$ is equilateral, then its area is $\dfrac{3\sqrt{3}}{2}|z_1|^2$
D) If triangle formed by $z_1,z_2,z_3$ is equilateral, then $z_1+z_2+z_3=0$
Step-by-Step Solution
Key Concept: The determinant $=0$ with $a=b=c$: circulant with $a=b=c$ always gives $0$. Also $|z_1|=|z_2|=|z_3|$ means they lie on a circle. The condition implies the triangle is equilateral or degenerates. For equilateral: orthocentre = circumcentre = $(z_1+z_2+z_3)/3$... but option B says $z_1+z_2+z_3$ (not divided by 3).
A ✓, B ✓, C ✗, D ✓. Answer: A, B, D.
Correct Answer: ABD