JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • question_answer If \[{{a}^{2}}+{{b}^{2}}=1,\] then \[\frac{1+b+ia}{1+b-ia}=\]

    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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