If {a^2} + {b^2} = 1, then frac{{1 + b + ia}} | vedclass

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(C) Given that ${a^2} + {b^2} = 1$. Consider the expression $Z = \frac{{1 + b + ia}}{{1 + b - ia}}$. Multiply the numerator and denominator by the con

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$b + ia$

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(C) Given that ${a^2} + {b^2} = 1$. Consider the expression $Z = \frac{{1 + b + ia}}{{1 + b - ia}}$. Multiply the numerator and denominator by the conjugate of the denominator,$(1 + b + ia)$: $Z = \frac{{(1 + b + ia)(1 + b + ia)}}{{(1 + b - ia)(1 + b + ia)}}$ $Z = \frac{{(1 + b)^2 + (ia)^2 + 2ia(1 + b)}}{{(1 + b)^2 + a^2}}$ Since ${a^2} + {b^2} = 1$,we have ${a^2} = 1 - {b^2}$. $Z = \frac{{1 + 2b + {b^2} - {a^2} + 2ia(1 + b)}}{{1 + 2b + {b^2} + {a^2}}}$ Substitute ${a^2} + {b^2} = 1$ in the denominator: $Z = \frac{{1 + 2b + {b^2} - (1 - {b^2}) + 2ia(1 + b)}}{{1 + 2b + 1}}$ $Z = \frac{{2{b^2} + 2b + 2ia(1 + b)}}{{2 + 2b}}$ $Z = \frac{{2b(b + 1) + 2ia(1 + b)}}{{2(1 + b)}}$ $Z = \frac{{2(1 + b)(b + ia)}}{{2(1 + b)}} = b + ia$ Thus,the correct option is $C$.

If ${a^2} + {b^2} = 1,$ then $\frac{{1 + b + ia}}{{1 + b - ia}} = $

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