Explanation

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