Question

If the origin and two points represented by complex numbers $A$ and $B$ from vertices of an equilateral triangle, then $\frac{A}{B}+\frac{B}{A}$ is equal to

• A. $1$
• B. $-1$
• C. $2$
• D. None of these

Answer 1

The correct option is A $1$

Let $A=z_1,B=z_2$ and $C=z_3,$ where $A,B,C$ are vertices of equilateral triangle.

Given that third point $C$ is origin, so $z_3=0.$

Let $z_2-z_3=\alpha ,z_3-z_1=\beta ,z_1-z_2=\gamma$$\Rightarrow z_2=\alpha ,-z_1=\beta ,z_1-z_2=\gamma$

$\therefore \alpha +\beta +\gamma =z_2-z_1+z_1-z_2=0$$\Rightarrow \overline{\alpha }+\overline{\beta }+\overline{\gamma }=0$ ...(1)

Since the triangle is equilateral triangle,

$\therefore BC=CA=AB$$\Rightarrow \left|(z_2-0)\right|=|0-z_1|=|z_1-z_2|$

$\Rightarrow \left|\alpha \right|=\left|\beta \right|=\left|\gamma \right|$$\Rightarrow {\left|\alpha \right|}^2={\left|\beta \right|}^2={\left|\gamma \right|}^2$$\Rightarrow \alpha \overline{\alpha }=\beta \overline{\beta }=\gamma \overline{\gamma }=k$ (say)

$\therefore \overline{\alpha }=\frac{k}{\alpha },\overline{\beta }=\frac{k}{\beta },\overline{\gamma }=\frac{k}{\gamma }$

Substituting values of $\overline{\alpha },\overline{\beta }$ and $\overline{\gamma }$ in (1), we get

$\frac{k}{\alpha }+\frac{k}{\beta }+\frac{k}{\gamma }=0$$\Rightarrow \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=0$$\Rightarrow \frac{1}{z_2}+\frac{1}{-z_1}+\frac{1}{z_1-z_2}=0$

$\Rightarrow \frac{z_1(z_1-z_2)-z_2(z_1-z_2)+z_1{z}_2}{z_1{z}_2(z_1-z_2)}=0$$\Rightarrow z_1^2-z_1{z}_2-z_1{z}_2+z_2^2+z_1{z}_2=0$

$\Rightarrow z_1^2+z_2^2=z_1{z}_2$$\Rightarrow \frac{z_1^2}{z_1{z}_2}+\frac{z_2^2}{z_1{z}_2}=1$$\Rightarrow \frac{z_1}{z_2}+\frac{z_2}{z_1}=1$

$\Rightarrow \frac{A}{B}+\frac{B}{A}=1.$