Complex Numbers And Quadratic Equations question 448
The statement $ (a+ib)<(c+id) $ is not true for [RPET 2002]
Options:
A) $ a^{2}+b^{2}=0 $
B) $ b^{2}+c^{2}=0 $
C) $ a^{2}+c^{2}=0 $
D) $ b^{2}+d^{2}=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ a+ib < c+id $ is defined if and only if its imaginary parts must be equal to zero, i.e. $ b=d=0 $. So, $ b^{2}+d^{2}=0 $.
BETA
