Question

If $\dfrac{1-ix}{1+ix}=a-ib$ and ${{a}^{2}}+{{b}^{2}}=1$, where a, b belongs to R then x equals to
(a) $\dfrac{2a}{\left( 1+{{a}^{2}} \right)+{{b}^{2}}}$
(b) $\dfrac{2b}{\left( 1+{{a}^{2}} \right)+{{b}^{2}}}$
(c) $\dfrac{2a}{\left( 1+{{b}^{2}} \right)+{{a}^{2}}}$
(d) $\dfrac{2b}{\left( 1+{{b}^{2}} \right)+{{a}^{2}}}$

Answer 1

Hint: To solve this question, we will first rationalize the left hand side of the equation given in question. From here, we will find a and b in terms of x and then we will put the values a and b in the equation ${{a}^{2}}+{{b}^{2}}=1$.

Complete step-by-step answer:
Here in this question, we will rationalize the given complex function on the left hand side of the equation. To rationalize the equation, we will multiply both the numerator and denominator on the left hand side with the denominator. After rationalization, we get the following:

& \Rightarrow \dfrac{1-ix}{1+ix}=a-ib \\\ & \Rightarrow \dfrac{1-ix}{1+ix}\times \dfrac{1-ix}{1-ix}=a-ib \\\ \end{aligned}$$ $$\Rightarrow \dfrac{{{\left( 1-ix \right)}^{2}}}{\left( 1+ix \right)\left( 1-ix \right)}=a-ib$$ - (i) Here, we are going to use the following identities in equation (i): $$\begin{aligned} & \Rightarrow {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ & \Rightarrow \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \\\ \end{aligned}$$ Thus, we will get the following equation: $$\begin{aligned} & \Rightarrow \dfrac{{{1}^{2}}+{{\left( ix \right)}^{2}}-2\left( 1 \right)\left( ix \right)}{{{1}^{2}}-{{\left( ix \right)}^{2}}}=a-ib \\\ & \Rightarrow \dfrac{1+{{i}^{2}}{{x}^{2}}-2ix}{1-{{i}^{2}}{{x}^{2}}}=a-ib \\\ \end{aligned}$$ Here, we will use the identity $${{i}^{2}}=-1$$. After doing this, we will get the following: $$\Rightarrow \dfrac{1+\left( -1 \right){{x}^{2}}-2ix}{1-\left( -1 \right){{x}^{2}}}=a-ib$$ $$\Rightarrow \dfrac{1-{{x}^{2}}-2ix}{1+{{x}^{2}}}=a-ib$$ - (ii) Here, we will use the fact that the above equation will satisfy only when the real parts and imaginary parts of the equation will be separately equal. Real part of the above equation is = $$\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$$. Imaginary part of the above equation = $$\dfrac{-2x}{1+{{x}^{2}}}$$. Thus, we will get following: $$\Rightarrow a=\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$$ - (iii) $$\Rightarrow -b=\dfrac{-2x}{1+{{x}^{2}}}$$ $$\Rightarrow b=\dfrac{2x}{1+{{x}^{2}}}$$ - (iv) Now, we will add 1 on both sides of equation (iii). After doing this, we will get following equation:$$\begin{aligned} & \Rightarrow a+1=\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}+1 \\\ & \Rightarrow a+1=\dfrac{1-{{x}^{2}}+1+{{x}^{2}}}{1+{{x}^{2}}} \\\ \end{aligned}$$ $$\Rightarrow a+1=\dfrac{2}{1+{{x}^{2}}}$$ - (v) Now, we will square both sides of the equation. After squaring both sides, we will get following equation: $$\Rightarrow {{\left( a+1 \right)}^{2}}={{\left( \dfrac{2}{1+{{x}^{2}}} \right)}^{2}}$$ $$\Rightarrow {{\left( a+1 \right)}^{2}}=\dfrac{4}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$$ - (vi). Now we will square both sides of the equations. After squaring both sides, we will get following equation: $$\Rightarrow {{\left( b \right)}^{2}}={{\left( \dfrac{2x}{1+{{x}^{2}}} \right)}^{2}}$$ $$\Rightarrow {{b}^{2}}=\dfrac{4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$$ - (vii) Now we will add equations (vi) and (vii). After doing this, we will get: $$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{4+4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$$ $$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{4\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$$ $$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{4}{1+{{x}^{2}}}$$ - (viii) Now we will multiply (iv) with 2. Thus, we will get: $$\Rightarrow 2b=\dfrac{4x}{1+{{x}^{2}}}$$ $$\Rightarrow \dfrac{4x}{1+{{x}^{2}}}=\dfrac{2b}{x}$$ - (ix) Now we will substitute $$\dfrac{4x}{1+{{x}^{2}}}$$ from (ix) to (viii). Thus, we get: $$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{2b}{x}$$ Now, we will take reciprocal on both sides: $$\begin{aligned} & \Rightarrow \dfrac{1}{{{\left( 1+a \right)}^{2}}+{{b}^{2}}}=\dfrac{x}{2b} \\\ & \Rightarrow x=\dfrac{2b}{{{\left( 1+a \right)}^{2}}+{{b}^{2}}} \\\ \end{aligned}$$ Hence, (b) is correct. Note: We have separated the real and imaginary parts and make them equal because we are given that x is a real number. If it is not given that x is a real number then we cannot separate the real and imaginary parts and do comparison.