Question

Express the complex number ${{(1+2i)}^{-2}}$ 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 reverse the term and remove the minus term from the power and then we proceed.

Complete step-by-step solution:
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). ${{i}^{2}}$ = -1
ii). ${{i}^{3}}$ = -i
iii). ${{i}^{4}}$ = 1
With these in mind, let us proceed with the question
${{(1+2i)}^{-2}}$= ${{\left( \dfrac{1}{1+2\text{i}} \right)}^{2}}$,
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.
= $\left( \dfrac{1}{1+2\text{i}} \right)\left( \dfrac{1-2\text{i}}{1-2\text{i}} \right)$,
= $\dfrac{1-2\text{i}}{1-\left( -4 \right)}$,
 = $\dfrac{1-2\text{i}}{5}$,
Now, we solve the square term:
${{\left( \dfrac{1-2\text{i}}{5} \right)}^{2}}$
= $\dfrac{{{\left( 1-2\text{i} \right)}^{2}}}{25}$,
= $\dfrac{\left( 1-4-4\text{i} \right)}{25}$’
= $\dfrac{-3-4\text{i}}{25}$
So, ${{(1+2i)}^{-2}}$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 the elimination of the imaginary part in the denominator which makes the question more approachable.