Given that the equation ${{z}^{2}}+\left( p+iq \right)z+r+is=0$, where p, q, r, s are real and non-zero has a real root, then
(a) $pqr={{r}^{2}}+{{p}^{2}}s$
(b) $prs={{q}^{2}}+{{r}^{2}}p$
(c) $qrs={{p}^{2}}+{{s}^{2}}q$
(d) $pqs={{s}^{2}}+{{q}^{2}}r$

Hint: Assume a solution and we use real and imaginary parts to create more equations and find relation between them. There is a need to remember that if a complex number is equal to zero then its real part and imaginary part is to be equal.

Complete step-by-step solution -
Given expression in the question:
${{z}^{2}}+\left( p+iq \right)z+r+is=0..........(i)$
Given information,
P, q, r and s are real.
There is a non-zero root for the equation.
An equation containing a complex number is called a complex equation.
A number containing imaginary term in it is called as complex number
There can be a real number which may satisfy a complex equation.
Example $ix+x=0,x=0$ also satisfies this equation.
So, here for a given complex equation we can assume a real root which satisfies the expression properly.
Let us assume z = k satisfies the expression, we know “k” is a real number.
Substituting z = k in expression we get:
${{k}^{2}}+\left( p+iq \right)k+r+is=0$
By simplifying we get:
${{k}^{2}}+pk+iqk+r+is=0$
By grouping real and imaginary terms separately we get:
 $\left( {{k}^{2}}+pk+r \right)+iqk+is=0$
By taking i common in last two terms, we get:
$\left( {{k}^{2}}+pk+r \right)+i\left( qk+s \right)=0$
We know,
By basic complex numbers knowledge, we say:
If a + ib = 0 then a = 0 and b = 0 must be true
By above we get 2 cases.
So, by equating each term to 0 we get:
Case 1: Equating imaginary part to 0.
qk + s = 0
By subtracting s on both sides, we get:
qk = -s
By dividing with q on both sides, we get:
$k=\dfrac{-s}{q}........(A)$
Case 2: Equating real part to zero, we get:
${{k}^{2}}+pk+r=0$
By algebraic properties, we say:
If $a{{x}^{2}}+bx+c=0$ , then its solutions can be written as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
By substituting equation (A) here, we get:
${{\left( \dfrac{-s}{q} \right)}^{2}}+p\left( \dfrac{-s}{q} \right)+r=0$
By simplifying, we get
${{s}^{2}}-pqs+{{q}^{2}}r=0$
By adding pqs on both sides, we get:
${{s}^{2}}+{{q}^{2}}r=pqs$
Therefore $pqs={{s}^{2}}+{{q}^{2}}r$
Option (d) is correct.

Note: First we need to equate the imaginary part to zero as the equation there is first degree it can be easily solved and can be plucked into the second degree equation for further simplification.