Question

The conjugate of $\dfrac{1}{{3 + 4i}}$ is

Answer 1

Hint : We can see that the above number in the denominator is a complex number. As we know that a complex number is a number which can be expressed in the $a + bi$ form, where $a$ and $b$ are real numbers and $i$ is the imaginary number. It means it consists of both real and imaginary parts. Now any complex number can be converted into trigonometric form also known as polar form i.e. $z = \left| z \right|\left( {\cos \theta + i\sin \theta } \right)$ . In this question we will rationalize the given term to get the required value.

Complete step-by-step answer :
As per the given we have a complex number $\dfrac{1}{{3 + 4i}}$ .
The real part of the given complex number is $3$ and the imaginary part of the above complex number is $4$ .
We can rationalize it with the same expression in the denominator but with different signs in the middle.
So we have:
$\dfrac{1}{{3 + 4i}} \times \dfrac{{3 - 4i}}{{3 - 4i}} = \dfrac{{3 - 4i}}{{(3 + 4i)(3 - 4i)}}$ .
Now we know an algebraic expression which says that $(a + b)(a - b) = {a^2} - {b^2}$ . Here we have $a = 3,b = 4$ .
By putting this in the formula we can write
$\dfrac{{3 - 4i}}{{({3^2} - 4{i^2})}} = \dfrac{{3 - 4i}}{{9 - 16{i^2}}}$ .
We know that the value of $i = \sqrt { - 1}$ , upon solving we have
$\dfrac{{3 - 4i}}{{9 - 16( - 1)}} = \dfrac{{3 - 4i}}{{9 + 16}} = \dfrac{{3 - 4i}}{{25}}$ .
Hence the required answer is $\dfrac{1}{{25}}(3 - 4i)$ .
So, the correct answer is “ $\dfrac{1}{{25}}(3 - 4i)$ ”.

Note : We should be careful while calculating the values and in the square of the imaginary part we should note that the square of any negative number is always positive, the negative sign changes. In the above solution, the value of $i = \sqrt { - 1}$ , so when we squared it, gives us ${i^2} = \sqrt { - 1} \times \sqrt { - 1} = - 1$ . We should not get confused with the value and write the value of ${i^2} = 1$ , it is the wrong value.