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

Hint: 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.