A) 1
B) 2
C) \[b+ia\]
D) \[a+ib\]
Correct Answer: C
Solution :
Given that\[{{a}^{2}}+{{b}^{2}}=1\], therefore \[\frac{1+b+ia}{1+b-ia}=\frac{(1+b+ia)(1+b+ia)}{(1+b-ia)(1+b+ia)}\] \[=\frac{{{(1+b)}^{2}}-{{a}^{2}}+2ia(1+b)}{1+{{b}^{2}}+2b+{{a}^{2}}}\]\[=\frac{(1-{{a}^{2}})+2b+{{b}^{2}}+2ia(1+b)}{2(1+b)}\] \[=\frac{2{{b}^{2}}+2b+2ia(1+b)}{2\,(1+b)}=b+ia\] 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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