Question

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

Answer 1

Hint: The above numbers are complex number and simplified form of a complex number is its standard form which is $a + ib,\;{\text{where }}a{\text{ and }}b{\text{ }} \in {\text{R}}$. The above complex number can be converted into its standard form with a rationalization method that multiply and divide the given complex number with the conjugate of its denominator. Conjugate of a complex number $z = a + ib$ is given as $\overline z = a - ib$

Complete step by step answer:
 In order to simplify the given complex number $z = \dfrac{{2 + i}}{{1 + 2i}}$, we have to write it in standard form which represents the real and the imaginary part of a complex number separately. To write the given complex number in standard form we will use the rationalization method to convert it into standard form. In the rationalization method we will multiply and divide the given complex number with the conjugate of its denominator.

Denominator of $z = \dfrac{{2 + i}}{{1 + 2i}}\;{\text{is}}\;1 + 2i$
Now conjugate of a complex number $z = a + ib$ is given as $\overline z = a - ib$
So the conjugate of $1 + 2i$ will be equals to $1 - 2i$
Now multiplying and dividing the given complex number $z = \dfrac{{2 + i}}{{1 + 2i}}$ with conjugate of $1 + 2i$ that is $1 - 2i$
$
\Rightarrow z = \dfrac{{(2 + i)}}{{(1 + 2i)}} \times \dfrac{{(1 - 2i)}}{{(1 - 2i)}} \\
\Rightarrow z = \dfrac{{(2 + i)(1 - 2i)}}{{(1 + 2i)(1 - 2i)}} \\ $
Simplifying the parentheses of numerator and denominator with help of distributive property of multiplication, we will get
$
\Rightarrow z = \dfrac{{(2 + i)(1 - 2i)}}{{(1 + 2i)(1 - 2i)}} \\
\Rightarrow z = \dfrac{{2(1 - 2i) + i(1 - 2i)}}{{\left( {{1^2} - {{\left( {2i} \right)}^2}} \right)}}\;\;\;\;\left[ {\because (a + b)(a - b) = {a^2} - {b^2}} \right] \\
\Rightarrow z = \dfrac{{2 - 4i + i - 2{i^2}}}{{\left( {1 - 4{i^2}} \right)}} \\
\Rightarrow z = \dfrac{{2 - 3i - 2{i^2}}}{{\left( {1 - 4{i^2}} \right)}} \\ $

Now we know that in complex numbers value of ${i^2} = - 1$, substituting this and simplifying further
$
\Rightarrow z = \dfrac{{2 - 3i - 2{i^2}}}{{\left( {1 - 4{i^2}} \right)}} \\
\Rightarrow z = \dfrac{{2 - 3i - 2 \times ( - 1)}}{{\left( {1 - 4 \times ( - 1)} \right)}} \\
\Rightarrow z = \dfrac{{2 - 3i + 2}}{{\left( {1 + 4} \right)}} \\
\Rightarrow z = \dfrac{{4 - 3i}}{5} \\ $
So we get \[z = \dfrac{{4 - 3i}}{5}\], writing it in standard form that is in real and imaginary form.
\[z = \dfrac{{4 - 3i}}{5} = \dfrac{4}{5} - \dfrac{{3i}}{5}\]
Also written as $\operatorname{Re} (z) = \dfrac{4}{5}\;{\text{and}}\;\operatorname{Im} (z) = \dfrac{{ - 3}}{5},\;where\;\operatorname{Re} (z)\;{\text{and}}\;\operatorname{Im} (z)$ are real and imaginary part of complex number respectively.

Note:A complex number is denoted by $z$ and symbol for its conjugate is given by $\overline z $. In a complex number when the real part of the number equals $0$ then the number is called purely imaginary and when its imaginary part equals $0$ then it is called purely real.