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

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 cirucmcenter of △ABC. Now, arg⁡z3z2=∠BOC=2∠BAC=2arg⁡z3−z1z2−z1=arg⁡z3−z1z2−z12 [∵∠BOC=2∠BAC] Hence, arg⁡z3z2=arg⁡z3−z1z2−z12 Also, centroid is z1+z2+z3/3. Since HG:GO≡2:1 (where z1+z2+z3 (by section formula). When triangle is equilateral centorid 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

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 cirucmcenter of △ABC. Now, arg⁡z3z2=∠BOC=2∠BAC=2arg⁡z3−z1z2−z1=arg⁡z3−z1z2−z12 [∵∠BOC=2∠BAC] Hence, arg⁡z3z2=arg⁡z3−z1z2−z12 Also, centroid is z1+z2+z3/3. Since HG:GO≡2:1 (where z1+z2+z3 (by section formula). When triangle is equilateral centorid 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 $z_{1}, z_{2}, z_{3}$ 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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$\arg (\frac{z_{3}}{z_{2}}) = \arg {(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}^{2}$

If triangle formed by $z_{1}, z_{2}, z_{3}$ is equilateral, then $z_{1} + z_{2} + z_{3} = 0$

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

Orthocenter of triangle formed by $z_{1}, z_{2}, z_{3}$ is $z_{1} + z_{2} + z_{3}$

answer is A, B, D.

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Detailed Solution

$|\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}| = 0 .$
$\begin{array}{l} \Rightarrow a^{3} + b^{3} + c^{3} - 3 a b c = 0 \\ \Rightarrow (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = 0 \\ \Rightarrow \frac{1}{2} (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0 \\ \Rightarrow (a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0 \\ \Rightarrow a = b = c [∵ a + b + c \neq 0, ∵ z_{1} \neq 0, ∵ |z_{1}| = a \neq 0 etc.] \end{array}$
$Hence, O A = O B = O C, where O is the origin and A, B, C are$
$the points representing z_{1}, z_{2} and z_{3}, respectively.$
$Therefore, O is cirucmcenter of △ A B C . Now,$
Question Image
$\begin{array}{l} \arg (\frac{z_{3}}{z_{2}}) = ∠ B O C \\ = 2 ∠ B A C = 2 \arg (\frac{z_{3} - z_{1}}{z_{2} - z_{1}}) \\ = \arg {(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}^{2} [∵ ∠ B O C = 2 ∠ B A C] \end{array}$
$Hence, \arg (\frac{z_{3}}{z_{2}}) = \arg {(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}^{2}$
$Also, centroid is (z_{1} + z_{2} + z_{3}) / 3 . Since H G : G O \equiv 2 : 1 (where$

$z_{1} + z_{2} + z_{3} (by section formula). When triangle is equilateral$
$centorid coincides with circumcenter; hence z_{1} + z_{2} + z_{3} = 0 .$
$Also, the area for equilateral triangle is (\sqrt{3} / 4) L^{2}, where L$ $is length of side. Since radius is |z_{1}|, L = \sqrt{3} |z_{1}|, the area is$ $(3 \sqrt{3} / 4) {|z_{1}|}^{2}$