Explanation

(b)
We have, $\left| \frac{\mathcal{z -}a}{\mathcal{z +}\overline{a}} \right| = 1$
$\Rightarrow \ \ \left| \mathcal{z -}a \right| = \left| \mathcal{z +}\overline{a} \right|\mathcal{\ \Rightarrow \ |z +}a\left. \ \right|^{2} = \left| \mathcal{z +}\overline{a} \right|^{2}$
$\Rightarrow \ \ \left( \mathcal{z -}a \right)\left( \overline{\mathcal{z -}a} \right) = \left( \mathcal{z +}\overline{a} \right)(\overline{\mathcal{z +}\overline{a}})$
$\Rightarrow \ \ \left( \mathcal{z -}a \right)\left( \overline{\mathcal{z}} - \overline{a} \right) = \left( \mathcal{z +}\overline{a} \right)(\overline{\mathcal{z}} + a)\ \ \ \lbrack\because\left( \overline{\overline{a}} \right) = a\rbrack$
$\mathcal{\Rightarrow \ \ z}\overline{\mathcal{z}}\mathcal{- z}\overline{a} - a\overline{\mathcal{z}} + a\overline{a}\mathcal{= z}\overline{\mathcal{z}}\mathcal{+ z}a + \overline{a}\overline{\mathcal{z}} + \overline{a}a$
$\mathcal{\Rightarrow \ z}a + \mathcal{z}\overline{a} + \overline{a}\overline{\mathcal{z}} + a\overline{\mathcal{z}} = 0$
$\Rightarrow \ \ \left( a + \overline{a} \right)\left( \mathcal{z +}\overline{\mathcal{z}} \right) = 0$
$\mathcal{\Rightarrow \ \ z +}\overline{\mathcal{z}} = 0\ \ \ \ \left\lbrack \because a + \overline{a} = 2\ Re(a) \neq 0 \right\rbrack$
$\Rightarrow \ \ 2Re\ \left( \mathcal{z} \right) = 0\ \Rightarrow 2x = 0$
$\Rightarrow \ \ x = 0\ \Rightarrow y$-axis