Question

Let $z_1$ and $z_2$ be two non-zero complex numbers such that $\frac{z_1}{z_2}+\frac{z_2}{z_1}=1$, then the origin and points represented by $z_1$ and $z_2$

• A. lie on a straight line
• B. form a right triangle
• C. form an equilateral triangle
• D. none of these

Answer 1

Answer: (C)

Let $z=\frac{z_1}{z_2}$, then $z+\frac{1}{z}=1 \Rightarrow z^2-z+1=0$ $\Rightarrow z=\frac{1 \pm \sqrt{3} i}{2} $
$ \Rightarrow \frac{z_1}{z_2}=\frac{1 \pm \sqrt{3} i}{2}$
If $z_1$ and $z_2$ are represented by $A$ and $B$ respectively and $O$ be the origin, then
$\frac{O A}{O B}=\frac{\left|z_1\right|}{\left|z_2\right|}=\left|\frac{1 \pm \sqrt{3} i}{2}\right|=\sqrt{\frac{1}{4}+\frac{3}{4}}=1 $
$\Rightarrow O A=O B $
$\text { Also, } \frac{A B}{O B}=\frac{\left|z_2-z_1\right|}{\left|z_2\right|}=\left|1-\frac{z_1}{z_2}\right| $
$ =\left|1-\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i\right)\right|=\left|\frac{1}{2} \mp \frac{\sqrt{3}}{2} i\right|=\sqrt{\frac{1}{4}+\frac{3}{4}}=1 $
$\Rightarrow A B =O B$
$\text { Thus, } O A =O B=A B .$
$\therefore \triangle O A B$ is an equilateral triangle.