Explanation

(a)
Here, $i\ \left\{ \log{\left( \frac{x - i}{x + i} \right)\ } \right\} –\pi + 2\tan^{- 1}{x = k\ (say)}$
$\therefore\ \log{\left( \frac{x + i}{x - i} \right) = i(k + \pi - 2\tan^{- 1}{x)}}$
or $\frac{x + i}{x - i} = e^{i\theta},$ where $\theta = k + \pi - 2\tan^{- 1}x$
$\Rightarrow \ x + i = (x\cos{\theta + \sin{\theta) + i(x\sin{\theta - \cos\theta}}})$
$\Rightarrow \ x = x\cos{\theta + \sin{\theta\ and\ 1 = x\sin{\theta - \cos\theta}}}$
$\Rightarrow \ x = \cot{\frac{\theta}{2}\ \ \Rightarrow \ \theta = 2\cot^{- 1}x}$
or $k + \pi - 2\tan^{- 1}{x = 2\cot^{- 1}x}$
$\Rightarrow \ \ k + \pi = 2(\cot^{- 1}{x + {tan^{- 1}}{x) = 2\left( \frac{\pi}{2} \right)}}$
$\Rightarrow \ \ k + \pi = \pi\ \ or\ \ k = 0$