
The value of \[{(1 + i)^8} + {(1 - i)^8}\]is
A. 16
B. -16
C. 32
D. -32
Answer
161.1k+ views
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.
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.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
