Explanation

(c)
Let $\mathcal{z -}1 = re^{i\alpha}$
$\therefore\ (x - 1) + iy = r(\cos{\alpha + i\sin{\alpha)}}$
$\therefore\ r^{2} = (x - 1)^{2} + y^{2}$
and $\tan{\alpha = \frac{y}{x - 1}}$
The expression
$\frac{\mathcal{z -}1}{e^{i\theta}} + \frac{e^{i\theta}}{\mathcal{z -}1} = re^{i(\alpha - \theta)} + \frac{1}{r}e^{- i(\alpha - \theta)}$
Which is given as real
$\therefore\ \ r\sin{(\alpha -}\theta) - \frac{1}{r}\sin{(\alpha - \theta) = 0}$
$\Rightarrow \ \ r - \frac{1}{r} = 0\ \ \Rightarrow \ r^{2} = 1$
$\Rightarrow \ \ (x - 1)^{2} + y^{2} = 1$
Which is a circle with centre (1, 0) and radius 1