Question

Express the complex number $\dfrac{1}{{{\left( 2+i \right)}^{2}}}$ in the standard form of a + ib:

Answer 1

Hint: First we will solve the denominator separately and write in the form of x + iy, then we will rationalize the denominator by multiplying x – iy in both numerator and denominator and then we have to rearrange some terms to make it in the form of a + ib.

Complete step-by-step answer:
The formula for ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we are going to use this formula for calculating the value of ${{z}^{2}}$, where z can any complex number.
Another formula that we are going to use is ${{i}^{2}}=-1$ ,
Let’s first solve denominator,
$\Rightarrow {{\left( 2+i \right)}^{2}}$
Now we will use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to expand,
$\Rightarrow \left( {{2}^{2}}+2\times 2i+{{i}^{2}} \right)$
Now we know that ${{i}^{2}}=-1$ , using this we get,
$\begin{aligned} & \Rightarrow \left( 4+4i-1 \right) \\\ & \Rightarrow \left( 3+4i \right) \\\ \end{aligned}$
After solving the denominator we get,
${{\left( 2+i \right)}^{2}}=\left( 3+4i \right)$
Now we will rationalize the denominator by multiplying (3 – 4i) in both numerator and denominator.
We are multiplying by (3 – 4i) because whenever we multiply a complex number by it’s conjugate we get a real number and that is our objective here.
Rationalizing the denominator means multiplying it with some numbers to make it into an integer.
$\dfrac{1}{{{\left( 2+i \right)}^{2}}}=\dfrac{1}{3+4i}$
Now multiplying (3 – 4i) in numerator and denominator we get,
$\begin{aligned} & \Rightarrow \dfrac{1}{{{\left( 2+i \right)}^{2}}}=\dfrac{\left( 3-4i \right)}{(3+4i)\left( 3-4i \right)} \\\ & \Rightarrow \dfrac{1}{{{\left( 2+i \right)}^{2}}}=\dfrac{\left( 3-4i \right)}{\left( 9-12i+12i-16{{i}^{2}} \right)} \\\ \end{aligned}$
Now we know that ${{i}^{2}}=-1$ , using this we get,
$\begin{aligned} & \Rightarrow \dfrac{1}{{{\left( 2+i \right)}^{2}}}=\dfrac{\left( 3-4i \right)}{25} \\\ & \Rightarrow \dfrac{1}{{{\left( 2+i \right)}^{2}}}=\dfrac{3}{25}+\left( \dfrac{-4}{25} \right)i \\\ \end{aligned}$
So, we converted the given equation in the form of a + ib, and now we will compare and find the value of a and b.
By comparing the value of a = $\dfrac{3}{25}$ and the value of b = $\dfrac{-4}{25}$ .
Hence, from this we can conclude that our given expression has been converted into the specified form.

Note: There are some concepts that one needs to understand this question that is the value of i is $\sqrt{-1}$ and ${{i}^{2}}=-1$, ${{i}^{3}}=-i$ and ${{i}^{4}}=1$, after that it repeats itself. And one more important concept is the conjugate of a complex number, the conjugate of ( a + ib ) is ( a – ib ), all these are required to solve this question.