If a^2 +b^2 =1 then, ((1+b+ia)/(1+b−ia))=?


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Question Number 167911 by MathsFan last updated on 29/Mar/22

$I f a^{2} + b^{2} = 1$ $t h e n, \frac{1 + b + i a}{1 + b - i a} = ?$
	Answered by MJS_new last updated on 29/Mar/22

	$\frac{b + 1 + a i}{b + 1 - a i} = \frac{{(b + 1 + a i)}^{2}}{(b + 1 - a i) (b + 1 + a i)} =$ $= \frac{- a^{2} + b^{2} + 2 b + 1 + 2 a (b + 1) i}{a^{2} + b^{2} + 2 b + 1} =$ $[b^{2} = 1 - a^{2}]$ $= \frac{- 2 a^{2} + 2 b + 2 + 2 a (b + 1) i}{2 b + 2} =$ $= \frac{- a^{2} + b + 1}{b + 1} + a i =$ $[a^{2} = 1 - b^{2}]$ $= \frac{b^{2} + b}{b + 1} + a i =$ $= b + a i$
	Commented by MathsFan last updated on 29/Mar/22

	$t h a n k s$
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