Explanation

(b)
Given, $\arg{\left( \frac{z - 1}{z + 1} \right) = \frac{\pi}{3}}$
Let $z = x + iy$
$\therefore\ \ \ \ \ \frac{z - 1}{z + 1} = \frac{x + iy - 1}{x + iy + 1} \times \frac{(x + 1) - iy}{(x + 1) - iy}$
$= \frac{x^{2} + y^{2} - 1 + 2iy}{(x + 1)^{2} + y^{2}}$
$\therefore\ \ \arg{\left( \frac{z - 1}{z + 1} \right) = \tan^{- 1}{\frac{2y}{x^{2} + y^{2} - 1} = \frac{\pi}{3}}}$
$\Rightarrow \ \ \ \ \ \ \frac{2y}{x^{2} + y^{2} - 1} = \sqrt{3}$
$\Rightarrow \ \ \ x^{2} + y^{2} - \frac{2}{\sqrt{3}}y - 1 = 0$
Which is the equation of a circle.