Question

Let $z_1,z_2,z_3$ be three non-zero complex numbers such that $z_2\ne 1,a=|z_1|,b=|z_2|$ and $c=|z_3|$. Let $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$. Then which of the following options is (are) CORRECT?

• A. $\arg \left(\frac{z_3}{z_2}\right)=\arg {\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2$
• 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 $\frac{3\sqrt{3}}{2}|z_1{|}^2$
• D. If triangle formed by $z_1,z_2,z_3$ is equilateral, then $z_1+z_2+z_3=0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\arg \left(\frac{z_3}{z_2}\right)=\arg {\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2$
B Orthocentre of triangle formed by$z_1,z_2,z_3$ is $z_1+z_2+z_3$
D If triangle formed by $z_1,z_2,z_3$ is equilateral, then $z_1+z_2+z_3=0$

Given, $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$
$\Rightarrow a^3+b^2+c^3-3abc=0$
$\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
$\Rightarrow \frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$
$\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$
$\Rightarrow a=b=c$
$[\because a+b+c\ne 0$ as $z_1,z_2,z_3$ are non-zero complex numbers, so $|z_i|\ne 0]$


Hence, $OA=OB=OC$, where $O$ is the origin and $A,B,C$ are the points representing $z_1,z_2$ and $z_3$, respectively. Therefore, $O$ is circumcentre of $△ABC.$

$\arg \left(\frac{z_3}{z_2}\right)=\angle BOC$
$=2\angle BAC(\because \angle BOC=2\angle BAC)$
$=2\arg \left(\frac{z_3-z_1}{z_2-z_1}\right)$
$=\arg {\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2$
$\Rightarrow \arg \left(\frac{z_3}{z_2}\right)=\arg {\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2$

Also, centroid is $\frac{(z_1+z_2+z_3)}{3}$.
Since, $HG:GO\equiv 2:1$ (where $H$ is orthocentre and $G$ is centroid), then orthocentre is $z_1+z_2+z_3$ (by section formula).
When triangle is equilateral, centroid coincides with the circumcentre, hence $z_1+z_2+z_3=0.$

Also, the area of the equilateral triangle is $\left(\frac{\sqrt{3}}{4}\right)L^2$, where $L$ is the length of side.
Since, radius is $|z_1|\Rightarrow L=\sqrt{3}|z_1|$
Hence, area is $\left(\frac{3\sqrt{3}}{4}\right)|z_1{|}^2$