Complex Numbers And Quadratic Equations question 416
Question: If $ a^{2}+b^{2}=1, $ then $ \frac{1+b+ia}{1+b-ia}= $
Options:
A) 1
B) 2
C) $ b+ia $
D) $ a+ib $
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ a^{2}+b^{2}=1 $ , therefore $ \frac{1+b+ia}{1+b-ia}=\frac{(1+b+ia)(1+b+ia)}{(1+b-ia)(1+b+ia)} $ $ =\frac{{{(1+b)}^{2}}-a^{2}+2ia(1+b)}{1+b^{2}+2b+a^{2}} $ $ =\frac{(1-a^{2})+2b+b^{2}+2ia(1+b)}{2(1+b)} $ $ =\frac{2b^{2}+2b+2ia(1+b)}{2(1+b)}=b+ia $ Trick: Put $ a=0,b=1 $ , $ \frac{1+b+ia}{1+b-ia}=\frac{1+1+0}{1+1-0}=1 $ But options A and C give 1. So again put $ a=1,b=0,\frac{1+b+ia}{1+b-ia}=\frac{1+i}{1-i}=i $ . Which gives © only.
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