Question

The value of \[{(1 + i)^8} + {(1 - i)^8}\]is
A. 16
B. -16
C. 32
D. -32

Answer 1

Hint: Here, \[i\] is known as iota which is used in complex numbers as an imaginary number. Its value is equal to \[\sqrt { - 1} \].
To solve this question, we use the algebraic formula for the square of two functions. After using it we can easily find out the result.

Formula used : Here, we use the formula shown below,
1.\[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
2. \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]

Complete step-by-step solution:
We are given that, \[{(1 + i)^8} + {(1 - i)^8}\]
Firstly, let’s find out the value of \[{\left( {1 + i} \right)^8}\].
To find the value of \[{\left( {1 + i} \right)^8}\] we rewrite the \[{\left( {1 + i} \right)^8}\] in the form of \[{(a + b)^2}\], we get
\[{\left( {1 + i} \right)^8} = {\left[ {{{\left( {1 + i} \right)}^2}} \right]^4}\] ……(1)
Now, we will apply the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] in equation (1) to simplify the expression, we get
\[{\left[ {{{\left( {1 + i} \right)}^2}} \right]^4} = {\left( {1 + {i^2} + 2i} \right)^4}\]
As we know that, \[{i^2} = - 1\].
So, we will substitute this value in the above expression and simplify this, we get
\[\begin{array}{l}{\left[ {{{\left( {1 + i} \right)}^2}} \right]^4} = {\left( {1 - 1 + 2i} \right)^4}\\{\left[ {{{\left( {1 + i} \right)}^2}} \right]^4} = {\left( {2i} \right)^4}\end{array}\]
Further, we will apply the exponent rule that is \[{a^n} = a \times a \times a \times ...... \times a\] and simplify this, we get
\[\begin{array}{l}{\left[ {{{\left( {1 + i} \right)}^2}} \right]^4} = 2i \times 2i \times 2i \times 2i\\{\left[ {{{\left( {1 + i} \right)}^2}} \right]^4} = {2^4} \times {i^4}\\{\left[ {{{\left( {1 + i} \right)}^2}} \right]^4} = 16{i^4}\end{array}\]
As we know, \[{i^2}\] is equal to \[ - 1\] then \[{i^4} = {i^2} \times {i^2}\] that is equal to \[1\].
Furthermore, we will substitute this value in above expression, we get
\[\begin{array}{l}{\left[ {{{\left( {1 + i} \right)}^2}} \right]^4} = 16 \times 1\\{\left[ {{{\left( {1 + i} \right)}^2}} \right]^4} = 16\end{array}\]
Now, we will substitute this value in equation (1), we get
\[{\left( {1 + i} \right)^8} = 16\] ……. (2)
Similarly, we will find \[{\left( {1 - i} \right)^8}\].
To find the value of \[{\left( {1 - i} \right)^8}\] we rewrite the \[{\left( {1 - i} \right)^8}\] in the form of \[{(a - b)^2}\], we get
\[{\left( {1 - i} \right)^8} = {\left[ {{{\left( {1 - i} \right)}^2}} \right]^4}\] ……(3)
Now, we will apply the formula \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] in equation (3) to simplify the expression, we get
\[{\left[ {{{\left( {1 - i} \right)}^2}} \right]^4} = {\left( {1 + {i^2} - 2i} \right)^4}\]
As we know that, \[{i^2} = - 1\].
So, we will substitute this value in the above expression and simplify this, we get
\[\begin{array}{l}{\left[ {{{\left( {1 - i} \right)}^2}} \right]^4} = {\left( {1 - 1 - 2i} \right)^4}\\{\left[ {{{\left( {1 - i} \right)}^2}} \right]^4} = {\left( { - 2i} \right)^4}\end{array}\]
Further, we will apply the exponent rule that is \[{a^n} = a \times a \times a \times ...... \times a\] and simplify this, we get
\[\begin{array}{l}{\left[ {{{\left( {1 - i} \right)}^2}} \right]^4} = \left( { - 2i} \right) \times \left( { - 2i} \right) \times \left( { - 2i} \right) \times \left( { - 2i} \right)\\{\left[ {{{\left( {1 - i} \right)}^2}} \right]^4} = {\left( { - 2} \right)^4} \times {i^4}\\{\left[ {{{\left( {1 - i} \right)}^2}} \right]^4} = 16{i^4}\end{array}\]
As we know, \[{i^2}\] is equal to \[ - 1\] then \[{i^4} = {i^2} \times {i^2}\] that is equal to \[1\].
Furthermore, we will substitute this value in above expression, we get
\[\begin{array}{l}{\left[ {{{\left( {1 - i} \right)}^2}} \right]^4} = 16 \times 1\\{\left[ {{{\left( {1 - i} \right)}^2}} \right]^4} = 16\end{array}\]
Now, we will substitute this value in equation (3), we get
\[{\left( {1 - i} \right)^8} = 16\] (4)
Now, we will substitute the value of equation (2) and equation (4) in given equation, we get
\[\begin{array}{l}{(1 + i)^8} + {(1 - i)^8} = 16 + 16\\{(1 + i)^8} + {(1 - i)^8} = 32\end{array}\]

Hence, option (C) is the correct answer.

Note : In these types of questions, we should remember the values of iota because there is a lot of confusion between the odd and even powers of iota. Here we should also remember the algebraic properties of a quadratic equation.