Let z1, z2, z3 be the three nonzero complex numbers such that z2≠1, a=z1, b=z2 and c=z3. Let abcbcacab=0. Then

Given, abcbcacab=0⇒ a3+b3+c3−3abc=0⇒ (a+b+c)a2+b2+c2−ab−bc−ca=0⇒ 12(a+b+c)(a−b)2+(b−c)2+(c−a)2=0⇒ (a−b)2+(b−c)2+(c−a)2=0⇒ a=b=c∵a+b+c≠0,∵z1≠0,∴z1=a≠0 etc. Hence, OA = OB = OC, where O is the origin and A, B, C are the points representing z1, z2 and z3, respectively. Therefore, O is circumcenter of △ABC. Now,arg⁡z3z2=∠BOC (1)=2∠BAC=2arg⁡z3−z1z2−z1 (2)=arg⁡z3−z1z2−z12 [∵∠BOC=2∠BAC]Hence, arg⁡z3z2=arg⁡z3−z1z2−z12Also, centroid is (z1 + z2 + z3)/3. Since HG : GO ≡ 2 : 1 (where H is orthocenter and G is centroid), then orthocenter is z1 + z2 + z3 (by section formula). When triangle is equilateral centroid coincides with circumcenter; hence z1 + z2 + z3 = 0.Also, the area for equilateral triangle is (3/4)L2, where L is length of side. Since radius is z1, L=3z1, the area is (33/4)z12.

Let z1, z2, z3 be the three nonzero complex numbers such that z2≠1, a=z1, b=z2 and c=z3. Let abcbcacab=0. Then

Given, abcbcacab=0⇒ a3+b3+c3−3abc=0⇒ (a+b+c)a2+b2+c2−ab−bc−ca=0⇒ 12(a+b+c)(a−b)2+(b−c)2+(c−a)2=0⇒ (a−b)2+(b−c)2+(c−a)2=0⇒ a=b=c∵a+b+c≠0,∵z1≠0,∴z1=a≠0 etc. Hence, OA = OB = OC, where O is the origin and A, B, C are the points representing z1, z2 and z3, respectively. Therefore, O is circumcenter of △ABC. Now,arg⁡z3z2=∠BOC (1)=2∠BAC=2arg⁡z3−z1z2−z1 (2)=arg⁡z3−z1z2−z12 [∵∠BOC=2∠BAC]Hence, arg⁡z3z2=arg⁡z3−z1z2−z12Also, centroid is (z1 + z2 + z3)/3. Since HG : GO ≡ 2 : 1 (where H is orthocenter and G is centroid), then orthocenter is z1 + z2 + z3 (by section formula). When triangle is equilateral centroid coincides with circumcenter; hence z1 + z2 + z3 = 0.Also, the area for equilateral triangle is (3/4)L2, where L is length of side. Since radius is z1, L=3z1, the area is (33/4)z12.

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Let z₁, z₂, z₃ be the three nonzero complex numbers such that $z_{2} \neq 1, a = |z_{1}|, b = |z_{2}|$ and $c = |z_{3}|$ . Let $|\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}| = 0$ . Then

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if triangle formed by z₁, z₂, z₃ is equilateral, then its area is $\frac{3 \sqrt{3}}{2} {|z_{1}|}^{2}$

if triangle formed by z₁, z₂, z₃ is equilateral, then z₁ + z₂ + z₃ = 0

$\arg (\frac{z_{3}}{z_{2}}) = \arg {(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}^{2}$

orthocenter of triangle formed by z₁, z₂, z₃ is z₁ + z₂ + z₃

answer is A, B, D.

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Detailed Solution

Given, $|\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}| = 0$

$\begin{array}{l} \Rightarrow a^{3} + b^{3} + c^{3} - 3 abc = 0 \\ \Rightarrow (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca) = 0 \\ \Rightarrow \frac{1}{2} (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0 \\ \Rightarrow (a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0 \\ \Rightarrow a = b = c [∵ a + b + c \neq 0, ∵ z_{1} \neq 0, ∴ |z_{1}| = a \neq 0 etc.] \end{array}$

Hence, OA = OB = OC, where O is the origin and A, B, C are the points representing z₁, z₂and z₃, respectively. Therefore, O is circumcenter of $△ ABC$ . Now,

$\begin{matrix} \arg (\frac{z_{3}}{z_{2}}) = ∠ BOC (1) \\ = 2 ∠ BAC = 2 \arg (\frac{z_{3} - z_{1}}{z_{2} - z_{1}}) (2) \\ = \arg {(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}^{2} [∵ ∠ BOC = 2 ∠ BAC] \end{matrix}$

Hence, $\arg (\frac{z_{3}}{z_{2}}) = \arg {(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}^{2}$

Also, centroid is (z₁ + z₂ + z₃)/3. Since HG : GO $\equiv$ 2 : 1 (where H is orthocenter and G is centroid), then orthocenter is z₁ + z₂ + z₃ (by section formula). When triangle is equilateral centroid coincides with circumcenter; hence z₁ + z₂ + z₃ = 0.

Also, the area for equilateral triangle is $(\sqrt{3} / 4) L^{2}$ , where L is length of side. Since radius is $|z_{1}|, L = \sqrt{3} |z_{1}|$ , the area is $(3 \sqrt{3} / 4) {|z_{1}|}^{2}$ .