Question

The points representing the complex number z for which arg $\left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$ lie on

• A. a circle
• B. a straight line
• C. an ellipse
• D. a parabola

Answer 1

Answer: (A)

Let $z=x +i y$
$\therefore \frac{z-2}{z+2}=\frac{x+ i y-2}{x +i y+2}$
$=\frac{(x-2)+i y}{(x+2)+i y} \times \frac{(x+2)-i y}{(x+2)-i y}$
$=\frac{(x-2)(x+2)+i y(x+2)-i y(x-2)-i^{2} y^{2}}{(x+2)^{2}-(i y)^{2}}$
$=\frac{x^{2}-4+i x y+2 i y-i x y+2 i y +y^{2}}{(x+2)^{2}+y^{2}}$
$=\frac{\left(x^{2}+y^{2}-4\right)+4 i y}{(x+2)^{2}+y^{2}}=\frac{x^{2}+y^{2}-4}{\left(x^{2}+4+4 x +y^{2}\right)}$
$+\frac{4 i y}{x^{2}+4+4 x +y^{2}}$
$\therefore \arg \left(\frac{z-2}{z+2}\right)= \tan ^{-1} \frac{4 y}{x^{2}+4+4 x +y^{2}}$
$\times \frac{x^{2}+4+4 x+ y^{2}}{x^{2}+y^{2}-4}=\frac{\pi}{3}$ (given)
$\therefore \tan \frac{\pi}{3}=\frac{4 y}{x^{2}+y^{2}-4}$
$\sqrt{3}=\frac{4 y}{x^{2}+y^{2}-4}$
$x^{2}+y^{2}-4-\frac{4}{\sqrt{3}} y=0$, which represents $a$ circle.