Question

The statement, $a+ib$ is less than $c+id$ is true for

• A. $a^2+b^2=0$
• B. $b^2+c^2=0$
• C. $a^2+c^2=0$
• D. $b^2+d^2=0$

Answer 1

The correct option is D

$b^2+d^2=0$

Explanation for the correct option:

Given, $a+ib$and $c+id$ can only be compared through the real parts $a$ and $c$.

So, in order for $a+ib<c+id$ to be true, they must be the real number which is possible when their imaginary parts are $0$.

Thus, $b=d=0\Rightarrow b^2+d^2=0$.

Hence, option (D), $b^2+d^2=0$ is the correct answer.