Question

If $\dfrac{1-ix}{1+ix}=a-ib$ and ${{a}^{2}}+{{b}^{2}}=1$, where $a,b\in R$ and $i=\sqrt{-1}$ , then the value of x should be equal to
(a) $\dfrac{2a}{{{\left( 1+a \right)}^{2}}+{{b}^{2}}}$
(b) $\dfrac{2b}{{{\left( 1+a \right)}^{2}}+{{b}^{2}}}$
(c) $\dfrac{2a}{{{\left( 1+b \right)}^{2}}+{{b}^{2}}}$
(d) $\dfrac{2b}{{{\left( 1+b \right)}^{2}}+{{b}^{2}}}$

Answer 1

Hint:In this question, we are given a relation involving x, a and b and another relation involving a and b. We should try to solve these equations to get a single equation with real and imaginary parts. As we know that in a complex equation, the real and imaginary part on both sides should be separately equal, we can equate the real and imaginary parts to obtain x in terms of a and b.

Complete step-by-step answer:
In this question, the first relation is given as
$\dfrac{1-ix}{1+ix}=a-ib$
Multiplying the numerator and denominator by \[\left( 1-ix \right)\] in the LHS, we obtain
\[\dfrac{{{\left( 1-ix \right)}^{2}}}{\left( 1+ix \right)\left( 1-ix \right)}=a-ib.................(1.1)\]
Now, for any two numbers, m and n, we have
$\left( m+n \right)\left( m-n \right)={{m}^{2}}-{{n}^{2}}..............(1.2)$
Using equation (1.2) in the denominator of equation (1.1), we have
\[\dfrac{{{\left( 1-ix \right)}^{2}}}{{{1}^{2}}-{{\left( ix \right)}^{2}}}=\dfrac{{{\left( 1-ix \right)}^{2}}}{{{1}^{2}}-{{i}^{2}}{{x}^{2}}}=a-ib.................(1.3)\]
However, as $i=\sqrt{-1}$, ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$, putting this value in the denominator of equation (1.3), we obtain
\[\begin{align}
  & \dfrac{{{\left( 1-ix \right)}^{2}}}{{{1}^{2}}-\left( -1 \right){{x}^{2}}}=a-ib \\
 & \Rightarrow {{\left( 1-ix \right)}^{2}}=\left( 1+{{x}^{2}} \right)\left( a-ib \right).................(1.4) \\
\end{align}\]
Expanding the LHS and RHS of equation (1.4) and using the fact that for any two numbers m and n, ${{\left( m-n \right)}^{2}}={{m}^{2}}+{{n}^{2}}-2mn$, and using the value of ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$, we can rewrite equation (1.4) as
\[\begin{align}
  & {{1}^{2}}+{{i}^{2}}{{x}^{2}}-2\times 1\times ix=a\left( 1+{{x}^{2}} \right)+i\left( -b\left( 1+{{x}^{2}} \right) \right) \\
 & \Rightarrow 1-{{x}^{2}}-2ix=a\left( 1+{{x}^{2}} \right)-bi\left( 1+{{x}^{2}} \right).................(1.5) \\
\end{align}\]
Now, the real and imaginary parts in both sides should be equal in both sides of equation (1.5), therefore, we obtain
\[\begin{align}
  & 1-{{x}^{2}}=a\left( 1+{{x}^{2}} \right)\Rightarrow a=\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}....................(1.6) \\
 & -2x=-b\left( 1+{{x}^{2}} \right)\Rightarrow b=\dfrac{2x}{1+{{x}^{2}}}..........................(1.7) \\
\end{align}\]
Now, from equations (1.6) and (1.7), we obtain
\[\begin{align}
  & {{\left( 1+a \right)}^{2}}+{{b}^{2}}={{\left( 1+\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)}^{2}}+{{\left( \dfrac{2x}{1+{{x}^{2}}} \right)}^{2}} \\
 & ={{\left( \dfrac{1+{{x}^{2}}+1-{{x}^{2}}}{1+{{x}^{2}}} \right)}^{2}}+{{\left( \dfrac{2x}{1+{{x}^{2}}} \right)}^{2}}={{\left( \dfrac{2}{1+{{x}^{2}}} \right)}^{2}}+{{\left( \dfrac{2x}{1+{{x}^{2}}} \right)}^{2}} \\
 & =\dfrac{4+4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\dfrac{4\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\dfrac{4}{\left( 1+{{x}^{2}} \right)}.....................(1.8) \\
\end{align}\]
And $2b=2\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\dfrac{4x}{1+{{x}^{2}}}..................(1.9)$
Therefore, dividing equation (1.8) from equation (1.9) we obtain
$\dfrac{2b}{{{\left( 1+a \right)}^{2}}+{{b}^{2}}}=\dfrac{\dfrac{4x}{1+{{x}^{2}}}}{\dfrac{4}{1+{{x}^{2}}}}=x\Rightarrow x=\dfrac{2b}{{{\left( 1+a \right)}^{2}}+{{b}^{2}}}$
Which is exactly the expression given in option (b), thus we see that option (b) should be the correct answer to this question.

Note: We should note that in equation (1.8), we should be careful to take out the factor of 4 in the numerator so that the factor of $\left( 1+{{x}^{2}} \right)$ cancels out from the numerator and the denominator. Otherwise, directly expanding the terms would result in terms involving ${{x}^{4}}$ which will be difficult to solve.And also by observing the options we have to simplify it ,it's an easy way for approaching a solution in a shortcut method.