Question

Express the complex number $ {{(2+i)}^{-2}} $ in the standard form of (a + ib).

Answer 1

Hint: We have to rationalize the given equation in the question and then we compare the following to the general term of a complex number which is a + ib. First, we inverse the term and remove the minus term from the power and then we proceed.

Complete step-by-step answer:
Complex numbers are numbers which are represented on the imaginary plane. They are represented in the following number: a + ib, where a denotes the real part of the complex number and b denotes the imaginary part.
Some of the basic identities we need to remember before we proceed into the question are
 $ {{i}^{2}} $ = 1
 $ {{i}^{3}} $ = -i
 $ {{i}^{4}} $ = 1
With these in mind, let us proceed with the question:
 $ {{(2+i)}^{-2}} $ = $ \dfrac{1}{{{\left( 2+\text{i} \right)}^{2}}} $
Now, we rationalize the term inside the bracket which means multiplying the number with its conjugate. For example, if we have to rationalize a + ib, we multiply the term with a – ib.
= $ \left( \dfrac{1}{2+\text{i}} \right)\left( \dfrac{2-\text{i}}{2-\text{i}} \right) $ ,
= $ \dfrac{2-\text{i}}{{{2}^{2}}-{{\left( \text{i} \right)}^{2}}} $
= $ \dfrac{2-\text{i}}{5} $ .
This is the term inside the bracket. We square it and then solve it obtain the rationalized term
 $ {{\left( \dfrac{2-\text{i}}{5} \right)}^{2}} $
= $ \dfrac{{{\left( 2-\text{i} \right)}^{2}}}{25} $ ,
= $ \dfrac{4-1-4\text{i}}{25} $ ,
= $ \dfrac{3-4\text{i}}{25} $ .
So, $ {{(2+i)}^{-2}} $ in the form of a + ib solves down to $ \dfrac{3-4\text{i}}{25} $ .

Note: When we multiply with the conjugate it gives us a simplified solution. So, remember that we have to take the conjugate carefully as it leads to elimination of the imaginary part in the denominator which makes the question more approachable.