Question

Solve it ${(1 + i)^{ - 3}}$

Answer 1

Hint : An imaginary number is the complex number which can be written as the real number multiplied by the imaginary unit “i”. Here we will use the laws of the power and exponents and simplify placing the values of the cubes and squares of the imaginary number “i”

** Complete step-by-step answer** :
${(1 + i)^{ - 3}}$
Take the inverse of the given term using the law of negative exponent.
${(1 + i)^{ - 3}} = \dfrac{1}{{{{(1 + i)}^3}}}$
Expand the denominator by using the formula to find the cube of the sum of two terms.
${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$
${(1 + i)^{ - 3}} = \dfrac{1}{{{1^3} + {i^3} + 3(1)(i)(1 + i)}}$
Simplify the above equation by opening the brackets and the constant outside the bracket.
${(1 + i)^{ - 3}} = \dfrac{1}{{{1^3} + {i^3} + 3i(1 + i)}} \\\ {(1 + i)^{ - 3}} = \dfrac{1}{{{1^3} + {i^3} + 3i + 3{i^2}}} \\\$
Now, substitute the values for “i”
${i^2} = - 1{\text{ and }}{{\text{i}}^3} = - i$ in the above equation-
${(1 + i)^{ - 3}} = \dfrac{1}{{{1^3} + ( - i) + 3i + 3( - 1)}}$
Also, place the value of the one cube equal to one. Also remember the basics of the mathematical operations, when you open the bracket plus minus is minus and minus minus gives plus and plus plus gives plus.
${(1 + i)^{ - 3}} = \dfrac{1}{{1 - i + 3i - 3}}$
Make a pair of like terms and simplify.
${(1 + i)^{ - 3}} = \dfrac{1}{{\underline {1 - 3} - \underline {i + 3i} }}$
When you simplify addition or subtraction between the terms, use the identity of plus minus you have to do subtraction and sign of greater digit.
$\Rightarrow {(1 + i)^{ - 3}} = \dfrac{1}{{ - 2 + 2i}}$
Take common from the denominator of the fraction on the right hand side of the equation.
$\Rightarrow {(1 + i)^{ - 3}} = \dfrac{1}{{ - 2(1 - i)}}$
Now, use the conjugate of the given equation and multiply and divide it with the above equation.
$\Rightarrow {(1 + i)^{ - 3}} = \dfrac{1}{{ - 2(1 - i)}} \times \dfrac{{(1 + i)}}{{(1 + i)}}$
Simplify the above equation and replace it by using the difference of two squares. $(a + b)(a - b) = {a^2} - {b^2}$
$\Rightarrow {(1 + i)^{ - 3}} = \dfrac{{(1 + i)}}{{ - 2(1{}^2 - {i^2})}}$
Now, substitute the values for “i”
${i^2} = - 1{\text{ }}$ in the above equation-
$\Rightarrow {(1 + i)^{ - 3}} = \dfrac{{(1 + i)}}{{ - 2(1 - ( - 1))}}$
Open the bracket, use the identity minus minus plus.

$$\Rightarrow {(1 + i)^{ - 3}} = \dfrac{{(1 + i)}}{{ - 2(1 + 1)}} \\\ \Rightarrow {(1 + i)^{ - 3}} = \dfrac{{(1 + i)}}{{ - 2(2)}} \\\ \Rightarrow {(1 + i)^{ - 3}} = \dfrac{{(1 + i)}}{{ - 4}} \\\$$

Hence, this is the required solution.
So, the correct answer is “ $\dfrac{{(1 + i)}}{{ - 4}}$ ”.

Note : Conjugate of the complex number is the number with an equal real part but an imaginary part equal magnitude but opposite in sign. It is expressed in the form of $a + ib$ , where “a” and “b” are the real numbers of the complex number.