Question

Find conjugate of the conjugate of the complex number $\dfrac{{3 + i}}{{2 - i}}$
A) $1 + i$
B) $1 - i$
C) $\dfrac{{1 + i}}{5}$
D) $\dfrac{{1 - i}}{5}$

Answer 1

In order to find the conjugate of a complex number, first we should know what conjugate is. In conjugate of a complex number the sign of the coefficient of iota changes. For example, if $a + ib$ is a complex number, then for its conjugate the sign before iota changes, and we get: $a - ib$, which is our conjugate for $a + ib$, where $a$ and $b$ are the real numbers.

Complete step by step solution:
We are given a complex number $\dfrac{{3 + i}}{{2 - i}}$.
From the definition of conjugate, we know that the sign of the coefficient of iota changes for conjugate of a complex number.
For example, if $a + ib$ is a complex number, then for its conjugate the sign before iota changes, and we get: $a - ib$, which is our conjugate for $a + ib$, where $a$ and $b$ are the real numbers.
The conjugate of a complex number is represented by dash and is given by: $\overline {a + ib} = a - ib$.
We are conjugating the given complex number as $\overline {\left( {\dfrac{{3 + i}}{{2 - i}}} \right)}$.
From the properties of complex number, we know that $\overline {\left( {\dfrac{{a + ib}}{{a - ib}}} \right)}$ is written as $\dfrac{{\left( {\overline {a + ib} } \right)}}{{\left( {\overline {a - ib} } \right)}}$.
Similarly, applying this to our complex number, we get: $\overline {\left( {\dfrac{{3 + i}}{{2 - i}}} \right)} = \dfrac{{\left( {\overline {3 + i} } \right)}}{{\left( {\overline {2 - i} } \right)}}$.
Changing the sign of the iota to find the conjugate of the complex number and we get: $\dfrac{{\left( {\overline {3 + i} } \right)}}{{\left( {\overline {2 - i} } \right)}} = \dfrac{{3 - i}}{{2 + i}}$.
Therefore, the conjugate of $\dfrac{{3 + i}}{{2 - i}}$ is $\dfrac{{3 - i}}{{2 + i}}$.
But we can see that the conjugate can be further simplified.
So, rationalizing the denominator by multiplying $2 - i$ to the numerator and denominator of the complex number:
$\dfrac{{3 - i}}{{2 + i}} \times \dfrac{{2 - i}}{{2 - i}}$
Simplifying it further:
$\dfrac{{\left( {3 - i} \right)\left( {2 - i} \right)}}{{\left( {2 + i} \right)\left( {2 - i} \right)}}$
Opening the parenthesis and solving it further and we get:
$\dfrac{{\left( {3 - i} \right)\left( {2 - i} \right)}}{{\left( {2 + i} \right)\left( {2 - i} \right)}} \\\ \Rightarrow \dfrac{{6 - 2i - 3i + {i^2}}}{{4 - 2i + 2i - {i^2}}} \\\ \Rightarrow \dfrac{{6 - 5i + {i^2}}}{{4 - {i^2}}} \\\$
Since, we know that $\sqrt { - 1} = i$, so squaring both the sides:
${\left( {\sqrt { - 1} } \right)^2} = {i^2} \\\ \Rightarrow {i^2} = \left( { - 1} \right) \\\$
Substituting ${i^2} = \left( { - 1} \right)$in $\dfrac{{6 - 5i + {i^2}}}{{4 - {i^2}}}$
$\dfrac{{6 - 5i + {i^2}}}{{4 - {i^2}}} = \dfrac{{6 - 5i - 1}}{{4 - \left( { - 1} \right)}} = \dfrac{{5 - 5i}}{{4 + 1}} = \dfrac{{5 - 5i}}{5}$
Taking $5$common from the numerator of the above value and we get:
$\dfrac{{5 - 5i}}{5} = \dfrac{{5\left( {1 - i} \right)}}{5}$
Cancelling $5$ from the numerator and denominator from the above equation, and we get:
$\dfrac{{5\left( {1 - i} \right)}}{5} = \left( {1 - i} \right)$, after this we cannot simplify it further.
So, the conjugate of $\dfrac{{3 + i}}{{2 - i}}$ in simplest form is $1 - i$, which matches with the option B.

Therefore, the correct option is Option (B) that is $1 - i$.

Note:

For rationalizing the denominator of a fraction, we multiply the numerator and denominator of the fraction by the conjugate of the denominator.
We can also leave the conjugate value in terms of fraction also, if it’s not needed to simplify.