Explanation

(a)
We have, $\mathcal{z =}\frac{1 + ir}{1 + p}\ \ \ \therefore i\mathcal{z =}\frac{- r + iq}{1 + p}$
By componendo and dividendo
$\frac{1 + i\mathcal{z}}{1 - i\mathcal{z}} - \frac{1 + p - r + iq}{1 + p + r - iq}$
$\therefore\ \frac{p + iq}{1 + r} = \frac{1 + i\mathcal{z}}{1 - i\mathcal{z}}\ if\ \frac{p + iq}{1 + r} = \frac{1 + p - r + iq}{1 + p + r - iq}$
or $p(1 + p + r) + q^{2} + i\{ q(1 + p + r) - pq\}$
$= (1 + r)(1 + p - r) + iq(1 + r)$
$\Rightarrow \ \ p(1 + p + r) + q^{2} = (1 + r)(1 + p - r)$
and $q(1 + p + r) - pq = q(1 + r)$
[this is obviously true]
$\therefore$ The condition is
$\ p(1 + p + r) + q^{2} = (1 + r)(1 + p - r)$
or $p + p^{2} + pr + q^{2} = 1 + p - r + r + pr - r^{2}$
or $p^{2} + q^{2} + r^{2} = 1$