Question

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

Answer 1

Hint: In this question we have been given with the expression for which we have to find the value of $x$ given that ${{a}^{2}}+{{b}^{2}}=1$. We will solve this question by first rationalizing the denominator by multiplying both the numerator and denominator by $\left( 1-ix \right)$ and get the term in the form of $a-ib$. We will then find ${{\left( 1+a \right)}^{2}}+{{b}^{2}}$ and then rearrange the expression to get the value of $x$ to get the required solution.

Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow \dfrac{\left( 1-ix \right)}{\left( 1+ix \right)}=\left( a-ib \right)$
On rationalizing the denominator by multiplying by $\left( 1-ix \right)$, we get:
$\Rightarrow \dfrac{\left( 1-ix \right)\left( 1-ix \right)}{\left( 1+ix \right)\left( 1-ix \right)}=\left( a-ib \right)$
On multiplying the terms, we get:
$\Rightarrow \dfrac{{{\left( 1-ix \right)}^{2}}}{{{1}^{2}}-{{\left( ix \right)}^{2}}}=\left( a-ib \right)$
On simplifying, we get:
$\Rightarrow \dfrac{{{\left( 1-ix \right)}^{2}}}{1-{{i}^{2}}{{x}^{2}}}=\left( a-ib \right)$
Now we know that ${{i}^{2}}=-1$, we get:
$\Rightarrow \dfrac{{{\left( 1-ix \right)}^{2}}}{1+{{x}^{2}}}=\left( a-ib \right)$
On using the expansion formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
$\Rightarrow \dfrac{1-2ix+{{i}^{2}}{{x}^{2}}}{1+{{x}^{2}}}=\left( a-ib \right)$
Now we know that ${{i}^{2}}=-1$, we get:
$\Rightarrow \dfrac{1-2ix-{{x}^{2}}}{1+{{x}^{2}}}=\left( a-ib \right)$
On splitting the terms with the real part on one side and the complex part on another side, we get:
$\Rightarrow \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}-i\dfrac{2x}{1+{{x}^{2}}}=\left( a-ib \right)$
On comparing, we get $a=\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$ and $b=\dfrac{2x}{1+{{x}^{2}}}$
Now on doing $1+a$, we get:
$\Rightarrow 1+a=1+\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$
On taking the lowest common multiple, we get:
$\Rightarrow 1+a=\dfrac{1+{{x}^{2}}+1-{{x}^{2}}}{1+{{x}^{2}}}$
On simplifying, we get:
$\Rightarrow 1+a=\dfrac{2}{1+{{x}^{2}}}$
On squaring both the sides, we get:
$\Rightarrow {{\left( 1+a \right)}^{2}}={{\left( \dfrac{2}{1+{{x}^{2}}} \right)}^{2}}$
On simplifying, we get:
$\Rightarrow {{\left( 1+a \right)}^{2}}=\dfrac{4}{{{\left( 1+{{x}^{2}} \right)}^{2}}}\to \left( 1 \right)$
Now we have $b=\dfrac{2x}{1+{{x}^{2}}}$ and ${{b}^{2}}=\dfrac{4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}\to \left( 2 \right)$
Now on adding equation $\left( 1 \right)$ and $\left( 2 \right)$, we get:
$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{4}{{{\left( 1+{{x}^{2}} \right)}^{2}}}+\dfrac{4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
Since the denominator is same, we get:
$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{4+4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
On taking $4$ common, we get:
$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{4\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$
On simplifying, we get:
$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{4}{1+{{x}^{2}}}$
Now we have $2b=\dfrac{4x}{1+{{x}^{2}}}$
Therefore, we can write it as:
$\Rightarrow {{\left( 1+a \right)}^{2}}+{{b}^{2}}=\dfrac{2b}{x}$
On rearranging, we get:
$\Rightarrow x=\dfrac{2b}{{{\left( 1+a \right)}^{2}}+{{b}^{2}}}$, which is the required solution.

Note: In these types of questions, it is to be remembered that complex numbers should be represented in the form of $a-ib$ and the real part and the imaginary part should be split. The value of $i$ should be remembered which is that $i$ is the square root of negative $1$ therefore, it can be written as $i=\sqrt{-1}$, therefore we can use the property ${{i}^{2}}=-1$.