Question

If z be any complex number such that $|3z-2|+|3z+2|=4$. then locus of z is

• A. an ellipse
• B. a circle
• C. a line segment
• D. None of these

Answer 1

The correct option is C a line segment

Given $|3z-2|+|3z+2|=4$

$\Rightarrow |z-\frac{2}{3}|+|z+\frac{2}{3}|=\frac{4}{3}$

let $z=x+iy\Rightarrow z-\frac{2}{3}=(x-\frac{2}{3})+iy$

$|z-\frac{2}{3}|=\sqrt{{(x-\frac{2}{3})}^2+y^2}---\left(1\right)$

similarly, $|z+\frac{2}{3}|=\sqrt{{(x+\frac{2}{3})}^2+y^2}---\left(2\right)$

substituting $\left(1\right).\left(2\right)$ in original equation

$\Rightarrow \sqrt{{(x-\frac{2}{3})}^2+y^2}+\sqrt{{(x+\frac{2}{3})}^2+y^2}=\frac{4}{3}$

$\sqrt{{(x-\frac{2}{3})}^2+y^2}=\frac{4}{3}-\sqrt{{(x+\frac{2}{3})}^2+y^2}$

Squaring both sides

${(x-\frac{2}{3})}^2+y^2=\frac{16}{9}+{(x+\frac{2}{3})}^2+y^2-\frac{8}{3}\sqrt{{(x+\frac{2}{3})}^2+y^2}$

$\Rightarrow \sqrt{(x+{\frac{2}{3}}^2+y^2)}=\frac{2}{3}+x$

${(x+\frac{2}{3})}^2+y^2={(x+\frac{2}{3})}^2$

$\Rightarrow y=0$ which is a line segment

$\therefore$ option $C$ is correct.