 # If ${a^2} + {b^2} = 1$ then $\dfrac{{1 + b + ia}}{{1 + b - ia}}$ is equal toA. $1$ B. $2$ C. $b + ia$ D. $a + ib$ Verified
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Hint: A complex number is of the form $a +ib$ where $a$ and $b$ are real numbers and $i$ is an imaginary number. We will use the formula $(a - b)(a + b) = {a^2} - {b^2}$ and ${(a + b)^2} = {a^2} + {b^2} + 2ab$ to solve the given expression. Also, we know that $i = \sqrt { - 1}$. Here, first we need to multiply by $1 + b + ia$ and then, we will make the given expression in the form of $a$ and $b$ after.

Complete step by step answer:
We need to find the value of the given expression as below,
$\dfrac{{1 + b + ia}}{{1 + b - ia}}$
Multiply the numerator and denominator by $1 + b + ia$, we will get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}} = \dfrac{{(1 + b) + (ia)}}{{(1 + b) - (ia)}} \times \dfrac{{(1 + b) + (ia)}}{{(1 + b) + (ia)}}$
We have used brackets so that we can take$a = 1 + b$ and $b = ia$(this is an assumption).
We will use the formula $(a - b)(a + b) = {a^2} - {b^2}$ and applying in the expression, we get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}}= \dfrac{{{{[(1 + b) + (ia)]}^2}}}{{{{(1 + b)}^2} - {{(ia)}^2}}}$
We will use the formula${(a + b)^2} = {a^2} + {b^2} + 2ab$ and applying it in the expression, we get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}} = \dfrac{{{{(1 + b)}^2} + {{(ia)}^2} + 2(1 + b)(ia)}}{{(1 + {b^2} + 2b) - {i^2}{a^2}}}$

Again we will use the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and applying it in the expression, we get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}} = \dfrac{{(1 + {b^2} + 2b) + {i^2}{a^2} + 2i(a + ab)}}{{1 + {b^2} + 2b - {i^2}{a^2}}}$
We know that$i = \sqrt { - 1} \Rightarrow {i^2} = - 1$.
Substituting the value of$i$in above expression, we will get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}} = \dfrac{{(1 + {b^2} + 2b) + ( - 1){a^2} + 2i(a + ab)}}{{1 + {b^2} + 2b - ( - 1){a^2}}}$
Removing the brackets, we get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}} = \dfrac{{1 + {b^2} + 2b - {a^2} + 2ia(1 + b)}}{{1 + {b^2} + 2b + {a^2}}}$
We are given that ${a^2} + {b^2} = 1$.

So, here we will substitute $1 = {a^2} + {b^2}$ in numerator and ${a^2} + {b^2} = 1$ in denominator, we will get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}} = \dfrac{{{a^2} + {b^2} + {b^2} + 2b - {a^2} + 2ia(1 + b)}}{{1 + 1 + 2{b^2}}}$
Simplify this above expression, we get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}}= \dfrac{{2{b^2} + 2b + 2ia(1 + b)}}{{2 + 2b}}$
$\Rightarrow \dfrac{{1 + b + ia}}{{1 + b - ia}} = \dfrac{{2b(b + 1) + 2ia(1 + b)}}{{2(1 + b)}}$
Dividing by $2$in both numerator and denominator, we get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}}= \dfrac{{b(b + 1) + ia(1 + b)}}{{(1 + b)}}$
Rearrange this above expression, we get,
$\dfrac{{1 + b + ia}}{{1 + b - ia}} = \dfrac{{b(1 + b) + ia(1 + b)}}{{(1 + b)}}$
$\Rightarrow \dfrac{{1 + b + ia}}{{1 + b - ia}}= \dfrac{{b(1 + b)}}{{(1 + b)}} + i\dfrac{{a(1 + b)}}{{(1 + b)}}$
Dividing by$(1 + b)$in both numerator and denominator, we get,
$\therefore \dfrac{{1 + b + ia}}{{1 + b - ia}}= b + ia$

Hence, if ${a^2} + {b^2} = 1$ then $\dfrac{{1 + b + ia}}{{1 + b - ia}}$ is $b + ib$.

Note: Complex Number is an algebraic expression including the factor $i = \sqrt { - 1}$. These numbers have two parts, one is called the real part and is denoted by Re (z) and the other is called the Imaginary Part called “iota”. Imaginary part is denoted by Im (z) for the complex number represented by 'z'. But either part can be $0$, so all Real Numbers and Imaginary Numbers are also Complex Numbers.