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The multiplication of a complex number and its conjugate is formulated as:

\[(a +ib)\] be the complex number and its conjugate is \[(a - ib)\]

\[ \Rightarrow (a - ib)(a - ib) = a(a - ib) + ib(a - ib)\]

\[ = {a^2} - iab + iab - {i^2}{b^2}\]

\[{a^2} + {b^2}\]

Given \[{a^2} + {b^2} = 1....(1)\]

We know, \[{a^2} + {b^2} = (b + ia)(b - ia)\]

Proof: On multiplying, we get:

\[(b + ia)(b - ia) = b(b - ia) + ia(b - ia)\]

\[ = {b^2} - bia + iab - {i^2}{a^2}\]

On cancelling \[ - bia\] and \[iab\], we get:

\[ = {b^2} - {i^2}{a^2}\]

Substituting \[{i^2} = - 1\], we get:

\[ \Rightarrow (b + ia)(b - ia) = {b^2} + {a^2}.........(2)\]

On using equation (2) in equation(1) we have

$\Rightarrow$ \[(b + ia)(b - ia) = 1\]

\[ \Rightarrow b - ia = \dfrac{1}{{b + ia}}..............(3)\]

Now we have to find the value of:

\[\dfrac{{(1 + b + ia)}}{{(1 + b + ia)}}..............(4)\]

Substituting \[b - ia = \dfrac{1}{{b + ia}}\] from (3) in (4), we have:

\[ \Rightarrow \dfrac{{1 + b + ia}}{{1 + \dfrac{1}{{b + ia}}}}\]

By taking LCM, we get:

\[ \Rightarrow \dfrac{{1 + b + ia}}{{\dfrac{{b + ia + 1}}{{b + ia}}}}\]

We know, \[\{ \because \dfrac{a}{{\dfrac{b}{c}}} = \dfrac{a}{b} \times c\} \]

\[ \Rightarrow \dfrac{{1 + b + ia}}{{1 + b + ia}} \times (b + ia)\]

On cancelling\[1 + b + ia\], we get

\[ \Rightarrow b + ia\]

Hence the required value is:

\[ \Rightarrow \dfrac{{1 + b + ia}}{{1 + b - ia}} = b + ia\]

In conjugate of a complex number only the imaginary part of that complex number changes its sign.