Let frac{1 - ix}{1 + ix} = a - ib and a^2 + b | vedclass

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(B) Given $\frac{1 - ix}{1 + ix} = a - ib$. Multiplying numerator and denominator by $(1 - ix)$,we get: $\frac{(1 - ix)^2}{1 + x^2} = a - ib$ $\frac{1

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$\frac{2b}{(1 + a)^2 + b^2}$

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(B) Given $\frac{1 - ix}{1 + ix} = a - ib$. Multiplying numerator and denominator by $(1 - ix)$,we get: $\frac{(1 - ix)^2}{1 + x^2} = a - ib$ $\frac{1 - x^2 - 2ix}{1 + x^2} = a - ib$ Equating real and imaginary parts: $a = \frac{1 - x^2}{1 + x^2}$ and $b = \frac{2x}{1 + x^2}$. We know that $a^2 + b^2 = 1$. Consider the expression $\frac{b}{1 + a} = \frac{\frac{2x}{1 + x^2}}{1 + \frac{1 - x^2}{1 + x^2}} = \frac{2x}{1 + x^2 + 1 - x^2} = \frac{2x}{2} = x$. Thus,$x = \frac{b}{1 + a}$. Since $a^2 + b^2 = 1$,we can write $1 = a^2 + b^2$. Substituting this into the denominator: $x = \frac{b}{1 + a} = \frac{2b}{2(1 + a)} = \frac{2b}{1 + 1 + 2a} = \frac{2b}{1 + (a^2 + b^2) + 2a} = \frac{2b}{(1 + a)^2 + b^2}$.

Let $\frac{1 - ix}{1 + ix} = a - ib$ and $a^2 + b^2 = 1$,where $a$ and $b$ are real,then $x = $

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