Question

Find the complex conjugate of $\dfrac{{3 + 2i}}{{1 - i}}$.

Answer 1

For the conjugate of a complex number the sign of the coefficient of iota changes. We will represent the conjugate of $\dfrac{{3 + 2i}}{{1 - i}}$ as $\overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)}$ then we will use the property of conjugate $\overline {\left( {\dfrac{{a + ib}}{{a - ib}}} \right)} = \dfrac{{\left( {\overline {a + ib} } \right)}}{{\left( {\overline {a - ib} } \right)}}$ to find the conjugate. We will then further simplify it by rationalising to find the simplest form.

Complete step by step answer:
As per the definition of conjugate, we know that for the conjugate of a complex number the sign of the coefficient of iota changes.
If $a + ib$ is a complex number then its conjugate will be given by changing sign before iota i.e., $a - ib$ is the conjugate of $a + ib$, where $a$ and $b$ are the real numbers. The conjugate of a complex number is represented by dash and is given by $\overline {a + ib} = a - ib$.
Given a complex number $\dfrac{{3 + 2i}}{{1 - i}}$ and we have to find the complex conjugate of this number.
Conjugate of $\dfrac{{3 + 2i}}{{1 - i}}$ is given as $\overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)}$.
Now using the property of complex number, we know that $\overline {\left( {\dfrac{{a + ib}}{{a - ib}}} \right)} = \dfrac{{\left( {\overline {a + ib} } \right)}}{{\left( {\overline {a - ib} } \right)}}$. Using this we get
$\Rightarrow \overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)} = \dfrac{{\left( {\overline {3 + 2i} } \right)}}{{\left( {\overline {1 - i} } \right)}}$
Changing the sign of iota, we get
$\Rightarrow \overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)} = \dfrac{{3 - 2i}}{{1 + i}}$
This conjugate can be further simplified.
Rationalizing the denominator by multiplying $\left( {1 - i} \right)$ to numerator and denominator of obtained conjugate, we get
$\Rightarrow \overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)} = \dfrac{{3 - 2i}}{{1 + i}} \times \dfrac{{1 - i}}{{1 - i}}$
$\Rightarrow \overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)} = \dfrac{{\left( {3 - 2i} \right)\left( {1 - i} \right)}}{{\left( {1 + i} \right)\left( {1 - i} \right)}}$
On simplifying, we get
$\Rightarrow \overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)} = \dfrac{{3 - 3i - 2i + 2{i^2}}}{{1 - i + i - {i^2}}}$
On further simplification, we get
$\Rightarrow \overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)} = \dfrac{{3 + 2{i^2} - 5i}}{{1 - {i^2}}}$
Putting the value of ${i^2} = - 1$, we get
$\Rightarrow \overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)} = \dfrac{{3 - 2 - 5i}}{{1 + 1}}$
On simplifying, we get
$\Rightarrow \overline {\left( {\dfrac{{3 + 2i}}{{1 - i}}} \right)} = \dfrac{1}{2}\left( {1 - 5i} \right)$
Therefore, the complex conjugate of $\dfrac{{3 + 2i}}{{1 - i}}$ is $\dfrac{{3 - 2i}}{{1 + i}}$ i.e., $\dfrac{1}{2}\left( {1 - 5i} \right)$.

Note:
We can solve this question using alternative methods as well. In this method first we will convert the given complex number to the simplest form of complex number by rationalisation. The simplest form is a+ib form and then we can find its conjugate.