• # question_answer The statement $(a+ib)<(c+id)$ is true for [RPET 2002] A) ${{a}^{2}}+{{b}^{2}}=0$ B) ${{b}^{2}}+{{c}^{2}}=0$ C) ${{a}^{2}}+{{c}^{2}}=0$ D) ${{b}^{2}}+{{d}^{2}}=0$

Correct Answer: D

Solution :

$a+ib<c+id,\,$ defined if and  only if its imaginary parts must be equal to zero, i.e. $b=d=0.$So,${{b}^{2}}+{{d}^{2}}=0$.

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