Explanation

(c)
Let $\mathcal{z -}1 = r(\cos{\theta + i\sin{\theta) =}}re^{i\theta}$
$\therefore\ \$Given expression$= re^{i\theta} \bullet e^{- i\alpha} + \frac{1}{re^{i\theta}}.e^{i\alpha}$
$= re^{i(\theta - \alpha)} + \frac{1}{r}e^{- i(\theta - \alpha)}$
Since, imaginary part of given expression is zero, we have
$r\sin{(\theta - \alpha) - \frac{1}{r}\sin{(\theta - \alpha) = 0}}$
$r^{2} - 1 = 0\ \Rightarrow \ r^{2} = 1\$
$\Rightarrow \ \ r = 1$
$\Rightarrow \ \ \left| \mathcal{z -}1 \right| = 1$
or $\sin{(\theta - \alpha) = 0\ \ \Rightarrow \ \theta - \alpha = 0}$
$\Rightarrow \ \ \theta = \alpha$
$\Rightarrow \ \arg{\left( \mathcal{z -}1 \right) = \alpha}$