Explanation

(d)
Given that,
$\mathcal{z}^{2} + (p + iq)\mathcal{z +}r + i\ s = 0$ …(i)
Let $\mathcal{z =}\alpha$ (where $\alpha$ is real) be a root of Eq. (i), then
$\alpha^{2} + ((p + iq)\alpha + r + i\ s = 0$ …(i)
$\Rightarrow \ \alpha^{2} + p\alpha + r + i(q\alpha + s) = 0$
On equating real and imaginary parts, we get
$\alpha^{2} + p\alpha + r = 0$ …(ii)
and $q\alpha + s = 0\ \Rightarrow \ \alpha = \frac{- s}{q}$
On putting the value of $\alpha$ in Eq.(ii), we get
$\left( \frac{- s}{q} \right)^{2} + p\left( \frac{- s}{q} \right) + r = 0$
$\Rightarrow \ \ s^{2} - pqs + q^{2}r = 0$
$\Rightarrow \ \ pqs = s^{2} + q^{2}r$