Question

How do you simplify $\dfrac{{3 - 2i}}{{3 + 2i}}$ ?

Answer 1

Hint:For solving this particular question, find the conjugate of the complex denominator. After multiplying the conjugate with the denominator and numerator , we get the real number in the denominator then use identities and simplify the given expression. At last, split the solution into its real and imaginary part.

Formula Used:
For solving the particular problem, we used two identities that are as follows.
$(a - b)(a + b) = {a^2} - {b^2}$
And
${(a - b)^2} = {a^2} - 2ab + {b^2}$
We also used ${i^2} = - 1$ .

Complete step by step answer:
We have to simplify the following expression,
$\dfrac{{3 - 2i}}{{3 + 2i}}$ ,
Now, we have to find the conjugate of the denominator .So, we can simplify it . Since the denominator that is $3 + 2i$ is a complex number therefore its conjugate is $3 - 2i$ . Now multiply the conjugate with the denominator as well as with the numerator , by multiplying the complex number with its conjugate gives a real number . Now, eliminate the complex number from the denominator of the given expression,
$ \Rightarrow \dfrac{{3 - 2i}}{{3 + 2i}} \times \dfrac{{3 - 2i}}{{3 - 2i}}$

Now using the identity $(a - b)(a + b) = {a^2} - {b^2}$ ,
We can write ,
$ \Rightarrow \dfrac{{{{(3 - 2i)}^2}}}{{{3^2} - {{(2i)}^2}}}$
Now using the identity ${(a - b)^2} = {a^2} - 2ab + {b^2}$ ,
We can write ,
$ \Rightarrow \dfrac{{9 + 4{{(i)}^2} - 2(3)(2i)}}{{9 - 4{{(i)}^2}}}$
Since we know that ${i^2} = - 1$ ,
$\Rightarrow \dfrac{{9 + 4( - 1) - 2(3)(2i)}}{{9 - 4( - 1)}} \\
\Rightarrow \dfrac{{5 - 12i}}{{13}} \\ $

$\therefore\dfrac{5}{{13}} - \dfrac{{12}}{{13}}i$ is the required solution.

Note: When we multiply complex numbers, multiplication is similar to the multiplication of two real numbers. In fact, multiplying two complex numbers is way similar to multiplying two binomials . Every term present in the first number must be multiplied with every term present in the other number. Any number which can be represented as $a + bi$ where $a$ and $b$ are real numbers then it is complex. And it exists on the complex plane.