Question

Let $z_1,z_2,z_3$ be the three nonzero complex numbers such that $z_1\ne 1,a=|z_1|,b=|z_2|$ and $c=|z_3|$. Let $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$
Then

• A. $arg\left(\frac{z_3}{z_2}\right)=arg(\frac{z_3-z_1}{z_2-z_1}{)}^2$
• B. Orthocenter 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 are 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)

$arg\left(\frac{z_3}{z_2}\right)=arg(\frac{z_3-z_1}{z_2-z_1}{)}^2$
Orthocenter of triangle formed by $z_1,z_2,z_3$ is $z_1+z_2+z_3$
If triangle formed by $z_1,z_2,z_3$ is equilateral, then $z_1+z_2+z_3=0$

Given,
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$
$\Rightarrow a^3+b^3+c^3-3abc=0$
$\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
$\Rightarrow \frac{1}{2}(a+b+c)\left]\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$
$\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$
$\Rightarrow a=b=c[\because a+b+c\ne 0,\because z_1\ne 0,\therefore a\ne 0$ etc.]
Hence, $OA=OB=OC$, where $O$ is the origin and $A,B,C$ are the point representing $z_1,z_2$ and $z_3$, respectively. Therefore, $O$ is circumcentre of $\angle ABC$.
Now,
$arg\left(\frac{z_3}{z_2}\right)=\angle BOC$ (i)
$=2\angle BAC$
$=2arg\left(\frac{z_3-z_1}{z_2-z_1}\right)$ (ii)
$=arg{\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2[\because \angle BOC=2\angle BAC]$
Hence, $arg\left(\frac{z_3}{z_2}\right)=arg{\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2$