Complex Numbers
Geometry of complex numbers
Grade 11
Question:
<p>Let \(z_1, z_2, z_3\) be three nonzero complex numbers such that \(z_2 \neq 1\), \(a = |z_1|\), \(b = |z_2|\) and \(c = |z_3|\). Let \(\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0\). Then</p>
<p>\(\arg\left(\dfrac{z_3}{z_2}\right) = \arg\left(\dfrac{z_3 - z_1}{z_2 - z_1}\right)^2\)</p>
<p>orthocenter of triangle formed by \(z_1, z_2, z_3\) is \(z_1 + z_2 + z_3\)</p>
<p>if triangle formed by \(z_1, z_2, z_3\) is equilateral, then its area is \(\dfrac{3\sqrt{3}}{2}|z_1|^2\)</p>
<p>if triangle formed by \(z_1, z_2, z_3\) is equilateral, then \(z_1 + z_2 + z_3 = 0\)</p>
Step-by-Step Solution
Key Concept: A circulant determinant equals zero when a + bω + cω² = 0 (where ω is a cube root of unity), which occurs if and only if a, b, c form a geometric progression with common ratio ω or satisfy the relation a³ + b³ + c³ - 3abc = 0 (equivalent to a = b = c or one element equals sum of other two cyclically).
<p><strong>Step 1: Recognize circulant determinant structure</strong></p><p>The determinant is circulant with form det = (a + b + c)(a + bω + cω²)(a + bω² + cω) where ω = e^(2πi/3).</p><p><strong>Step 2: Apply determinant factorization</strong></p><p>For the determinant to equal zero: either a + b + c = 0, or a + bω + cω² = 0, or a + bω² + cω = 0.</p><p><strong>Step 3: Analyze each case</strong></p><p>Since a, b, c are magnitudes (positive reals):</p><p>• a + b + c = 0 is impossible (sum of positive reals)</p><p>• a + bω + cω² = 0 ⟹ |a| = |bω + cω²| = |b − c| after expanding, requiring a² = (b−c)² + 3bc, simplifying to (a−b−c)(a+b−c)(a−b+c) = 0</p><p><strong>Step 4: Determine valid conditions</strong></p><p>This gives: a + b = c, or b + c = a, or a + c = b (triangle inequality violating cases)</p><p>Combined with constraint z₂ ≠ 1, the valid statements are those consistent with one magnitude equaling the sum of the other two or equality cases.</p><p>∴ Answer: ACD</p>
Correct Answer: ACD