If a linear inequality in complex number a+ib < c+id is meaningful if
A). $$a^{2}+b^{2}=0$$
B). $$b^{2}+c^{2}=0$$
C). $$a^{2}+c^{2}=0$$
D). $$b^{2}+d^{2}=0$$

Hint: In this question it is given that we have to find the condition for which the given condition a+ib < c+id is meaningful. So to find the solution we need to know any two complex numbers cannot be compared as there is no such concept. So by using this concept we have to solve the problem.

Complete step-by-step solution:
Here the given condition is a+ib < c+id.
As we know that Imaginary numbers cannot be compared.
So the condition a+ib < c+id to be meaningful, if and only if they are real numbers. Which is possible when their imaginary parts are zero.
i.e. b and d must be zero, b=0, d=0.
Therefore we can say that $$b^{2}+d^{2}=0$$.
Hence the correct option is option D.

Note: While solving this type of question you need to know that complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation $$i^{2}$$ = −1. Because no real number satisfies this equation, it is called an imaginary number. For the complex number a + bi, a is called the real part, and b is called the imaginary part. Also we can compare two complex numbers for equality. That is, we can assert or question if two complex numbers $$z_{1},z_{2}$$ are equal, i.e, $$z_{1}= z_{2}$$.
But there isn’t an ordering on complex numbers which follows all the rules we would expect of an ordering. So we can’t say $$z_{1} >z_{2}$$, $$z_{1} < z_{2}$$ or $$z_{1}\leq z_{2}$$ in sensible manner.