Question

Let $\left(\frac{1-ix}{1+ix}\right)=\left(a-ib\right)$ and $a^2+b^2=1$ where $a$ and $b$ are real, then $x$ equal to

Answer 1

Step 1: Rationalizing the denominator of left-hand side $x$

Given, $\left(\frac{1-ix}{1+ix}\right)=a-ib$

Rationalizing the denominator, we get

$$\begin{gathered} \left(\frac{1-ix}{1+ix}\right)\times \left(\frac{1-ix}{1-ix}\right)=a-ib \\ \Rightarrow \frac{{\left(1-ix\right)}^2}{\left(1+ix\right)\left(1-ix\right)}=a-ib \\ \Rightarrow \frac{1-2ix+i^2{x}^2}{1^2-{\left(1x\right)}^2}=a-ib\mathbf{}\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ \Rightarrow \frac{1-2ix-x^2}{1+x^2}=a-ib\left[\because i=\sqrt{-1},i^2=-1\right] \\ \Rightarrow \frac{1-x^2}{1+x^2}-i\frac{2x}{1+x^2}=a-ib \\ \Rightarrow a=\frac{1-x^2}{1+x^2},b=\frac{2x}{1+x^2} \end{gathered}$$

Now adding $1$to $a$ we get,

$$\begin{gathered} 1+a=1+\frac{1-x^2}{1+x^2} \\ \Rightarrow 1+a=\frac{1+x^2+1-x^2}{1+x^2} \\ \Rightarrow 1+a=\frac{2}{1+x^2} \\ \Rightarrow {\left(1+a\right)}^2=\frac{4}{{\left(1+x\right)}^2}\left[\because \mathrm{squaring}\mathrm{both}\mathrm{sides}\right] \end{gathered}$$

Similarly, $b^2=\frac{4x^2}{{\left(1+x^2\right)}^2}$

Now adding ${\left(1+a\right)}^2$ and $b^2$ we get,

$\begin{array}{rcl}{\left(1+a\right)}^2+b^2& =& \frac{4}{{\left(1+x^2\right)}^2}+\frac{4x^2}{{\left(1+x^2\right)}^2}\\ & =& \frac{4+4x^2}{{\left(1+x^2\right)}^2}\\ & =& \frac{4\left(1+x^2\right)}{{\left(1+x^2\right)}^2}\\ & =& \frac{4}{\left(1+x^2\right)}\end{array}$

Step 2: Finding the value of $x$

Now we know that, $b=\frac{2x}{1+x^2}$. Multiply both sides by $2$we get,

$$\begin{gathered} 2b=\frac{4x}{1+x^2} \\ \Rightarrow x=\frac{2b\left(1+x^2\right)}{4} \\ \Rightarrow x=\frac{2b}{{\displaystyle \frac{4}{\left(1+x^2\right)}}} \\ \Rightarrow x=\frac{2b}{{\left(1+a\right)}^2+b^2}\left[\because {\left(1+a\right)}^2+b^2=\frac{4}{1+x^2}\right] \end{gathered}$$

Hence, the value of $x$ equals to $\frac{2b}{{\left(1+a\right)}^2+b^2}$