Explanation

(c)
Given, $z = x + iy$
$\therefore\ \ \ \ \ \ \frac{z - 1}{z + 1} = \frac{x + iy - 1}{x + iy + 1}$
$\frac{(x - 1) + iy}{(x + 1) + iy} \times \frac{(x + 1) - iy}{(x + 1) - iy}$
$= \frac{x^{2} + y^{2} + 2iy - 1}{x^{2} + 1 + 2x + y^{2}}$
$\therefore\ \ \ \ \ \arg\left( \frac{z - 1}{z + 1} \right) = \tan^{- 1}\frac{2y}{x^{2} + y^{2} - 1}$
$\Rightarrow \ \ \ \ \ \tan^{- 1}{\frac{2y}{x^{2} + y^{2} - 1} = \frac{\pi}{4}}$ [given]
$\Rightarrow \ \ \ \ \frac{2y}{x^{2} + y^{2} - 1} = \tan{\frac{\pi}{4} = 1}$
$\Rightarrow \ \ \ \ \ \ x^{2} + y^{2} - 2y = 1$