Question

How do you simplify $\dfrac{{1 + 3i}}{{2 + 2i}}$.

Answer 1

Hint: This type of problem of complex numbers or imaginary numbers will be solved by removing the imaginary number $i$ as much as possible. Here we will multiply the numerator and the denominator by $(2 - 2i)$. Then we will get some term ${i^2}$ both in numerator and denominator which is a real number $ - 1$ . In this way, the given problem will be simplified.

Formula used: Used formulas:
$(a + b)(a - b) = {a^2} - {b^2}$
$i = \sqrt { - 1} $
$ \Rightarrow {i^2} = - 1$

Complete step-by-step solution:
We have;
$\dfrac{{1 + 3i}}{{2 + 2i}}$
Now we will try to remove $i$ from the denominator. For that, we have to find a term, which can remove $i$ from the denominator after multiplying the numerator and the denominator by that term.
Multiplying the numerator and the denominator by $(2 - 2i)$ we will get;
$ = \dfrac{{(1 + 3i)(2 - 2i)}}{{(2 + 2i)(2 - 2i)}}$
As we know $(a + b)(a - b) = {a^2} - {b^2}$ ;
Here $a = 2$ and $b = 2i$ we will get ;
$ = \dfrac{{2 - 2i + 6i - 6{i^2}}}{{4 - 4{i^2}}}$
We also know that ;
$i = \sqrt { - 1} $
$ \Rightarrow {i^2} = - 1$
Putting this value we will get ;
$ = \dfrac{{2 - 2i + 6i - 6( - 1)}}{{4 - 4( - 1)}}$
$ = \dfrac{{2 - 2i + 6i + 6}}{{4 + 4}}$
Adding all terms in numerator and denominator we will get ;
$ = \dfrac{{8 + 4i}}{8}$
Taking $8$ common from the numerator we will get;
$ = \dfrac{{8\left( {1 + \dfrac{1}{2}i} \right)}}{8}$
Hence $8$ will be vanished from the numerator and denominator we will get the final solution ;
$ = \dfrac{{1 + \dfrac{1}{2}i}}{1}$
$ = 1 + \dfrac{1}{2}i$

So the simplified form is $1 + \dfrac{1}{2}i$ where $1$ is called the real part and $\dfrac{1}{2}i$ is called the imaginary part.

Note: $\sqrt { - 1} $ is called the imaginary number and it is denoted by $i$ . To simplify this kind of problem we will try to transform $i$ in ${i^2}$ which is none other than a real number $ - 1$. By introducing $ - 1$ the summation or subtraction in between the numbers will be possible and then the problem will be simplified. Normally we can’t do the algebraic operations in real numbers and an imaginary number.