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Question: How do you simplify \(\dfrac{{1 + 3i}}{{2 + 2i}}\)....

How do you simplify 1+3i2+2i\dfrac{{1 + 3i}}{{2 + 2i}}.

Explanation

Solution

This type of problem of complex numbers or imaginary numbers will be solved by removing the imaginary number ii as much as possible. Here we will multiply the numerator and the denominator by (2−2i)(2 - 2i). Then we will get some term i2{i^2} both in numerator and denominator which is a real number −1 - 1 . In this way, the given problem will be simplified.

Formula used: Used formulas:
(a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}
i=−1i = \sqrt { - 1}
⇒i2=−1\Rightarrow {i^2} = - 1

Complete step-by-step solution:
We have;
1+3i2+2i\dfrac{{1 + 3i}}{{2 + 2i}}
Now we will try to remove ii from the denominator. For that, we have to find a term, which can remove ii from the denominator after multiplying the numerator and the denominator by that term.
Multiplying the numerator and the denominator by (2−2i)(2 - 2i) we will get;
=(1+3i)(2−2i)(2+2i)(2−2i)= \dfrac{{(1 + 3i)(2 - 2i)}}{{(2 + 2i)(2 - 2i)}}
As we know (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2} ;
Here a=2a = 2 and b=2ib = 2i we will get ;
=2−2i+6i−6i24−4i2= \dfrac{{2 - 2i + 6i - 6{i^2}}}{{4 - 4{i^2}}}
We also know that ;
i=−1i = \sqrt { - 1}
⇒i2=−1\Rightarrow {i^2} = - 1
Putting this value we will get ;
=2−2i+6i−6(−1)4−4(−1)= \dfrac{{2 - 2i + 6i - 6( - 1)}}{{4 - 4( - 1)}}
=2−2i+6i+64+4= \dfrac{{2 - 2i + 6i + 6}}{{4 + 4}}
Adding all terms in numerator and denominator we will get ;
=8+4i8= \dfrac{{8 + 4i}}{8}
Taking 88 common from the numerator we will get;
=8(1+12i)8= \dfrac{{8\left( {1 + \dfrac{1}{2}i} \right)}}{8}
Hence 88 will be vanished from the numerator and denominator we will get the final solution ;
=1+12i1= \dfrac{{1 + \dfrac{1}{2}i}}{1}
=1+12i= 1 + \dfrac{1}{2}i

So the simplified form is 1+12i1 + \dfrac{1}{2}i where 11 is called the real part and 12i\dfrac{1}{2}i is called the imaginary part.

Note: −1\sqrt { - 1} is called the imaginary number and it is denoted by ii . To simplify this kind of problem we will try to transform ii in i2{i^2} which is none other than a real number −1 - 1. By introducing −1 - 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.