Question

How do you simplify \[\dfrac{{3 + 2i}}{{2 + i}}\] ?

Answer 1

Hint:Iota (“i”) is known as the complex number whose value is minus root one, it was found by the mathematician to deal the negative sign under root, previously when this was not defined then if negative sign comes under root then there was no solution for that, but after this research complex terms can now easily be solved and tackle.

Complete step by step answer:
The mathematical expression here is a complex number. First, we will try to get rid of \[i\] from the denominator. For this, we will find the conjugate of the denominator. The denominator here is \[2 + i\] and its conjugate is \[2 - i\]. We will multiply this conjugate with the expression. We will multiply it both on the numerator and also in the denominator, and we will get:
\[\dfrac{{(3 + 2i) \cdot (2 - i)}}{{(2 + i) \cdot (2 - i)}}\]
Here, we will start solving and simplifying the expression. We know that \[i = \sqrt { - 1} \], then \[{i^2} = - 1\]. We will put the value of \[{i^2}\]in the expression, and we will get:
\[\dfrac{{6 - 3i + 4i - 2{i^2}}}{{4 - {i^2}}}\]
\[\Rightarrow\dfrac{{6 + i - 2( - 1)}}{{4 - ( - 1)}}\]
\[\Rightarrow\dfrac{{6 + i + 2}}{5}\]
\[\Rightarrow\dfrac{{8 + i}}{5}\]
Now, we will apply the standard form of a complex number. The standard form is:
\[a \pm bi\]
Here, \[a\]is the part where the real number lies. Now, we will put our answer according to the standard form of a complex number. The answer is:
\[\therefore \dfrac{8}{5} + \dfrac{1}{5}i\]
Therefore, this is our final answer.

Additional information:
Dealing with the complex equation you have to be careful only when you are dealing in higher degree equations because there the value of “iota” is given as for higher degree terms and accordingly the question needs to be solved.

Note:After the development of iota, research leads with the formulas associated and the properties like summation, subtraction, multiplication and division for the complex numbers. Graphs for complex numbers are also designed and the area under which the graph is drawn contains complex numbers only, but relation between complex and real numbers can be drawn.