Question

If $z_1,z_2,z_3$ are complex numbers such that $\left(\frac{2}{z_1}\right)=\left(\frac{1}{z_2}\right)+\left(\frac{1}{z_3}\right)$ and $\arg \left(\frac{z_3}{z_2}\right)\ne n\pi ,n\in I$, then the points $z_1,z_2,z_3$ and $O$(origin) will always lie on

• A. a circle
• B. a straight line
• C. An equilateral triangle with $O$ as centriod
• D. a rectangle

Answer 1

The correct option is A a circle

$\frac{2}{z_1}=\frac{1}{z_2}+\frac{1}{z_3}$
$\Rightarrow \frac{1}{z_1}-\frac{1}{z_2}=\frac{1}{z_3}-\frac{1}{z_1}$
$\Rightarrow \frac{z_2-z_1}{z_1{z}_2}=\frac{z_1-z_3}{z_3{z}_1}$
$\Rightarrow \frac{z_3-z_1}{z_2-z_1}=-\frac{z_3}{z_2}$

Now $\arg \left(\frac{z_3-z_1}{z_2-z_1}\right)=\arg (-\frac{z_3}{z_2})$
$\Rightarrow \arg \left(\frac{z_3-z_1}{z_2-z_1}\right)=\pi +\arg \left(\frac{z_3}{z_2}\right)\Rightarrow \arg \left(\frac{z_3-z_1}{z_2-z_1}\right)+\arg \left(\frac{z_2-0}{z_3-0}\right)=\pi$
Let $\alpha =\arg \left(\frac{z_3-z_1}{z_2-z_1}\right),\beta =\arg \left(\frac{z_2-0}{z_3-0}\right)$
then, $\alpha +\beta =\pi$
and possible diagram will be

Hence, the said points are concyclic.