Evaluate: $ {(1 + i)^6} + {(1 - i)^3} $
Answer
591.6k+ views
Hint: We will first expand the given expression using algebraic identities. Note that $ {i^2} $ is -1 and $ {i^3} $ is -1. Also $ {i^4} $ is equal to 1. We can use these results to simplify the expressions whenever possible.
Complete step-by-step answer:
The given expression involves the iota $ i = \sqrt { - 1} $ . From its definition we can see that the consecutive powers of iota are $ {i^2} = - 1 $ , $ {i^3} = - 1.i = - i $ , $ {i^4} = {( - 1)^2} = 1 $ . Also note that $ {i^5} = {i^4}.i = 1.i = i $ . So, the values of the next consecutive four powers repeat again. In general, $ {i^{4n}} = 1,{i^{4n + 1}} = i,{i^{4n + 2}} = - 1 $ and $ {i^{4n + 3}} = - i $ . We will use this observation while solving the given expression.
We can use algebraic identities to solve the given expression. We will evaluate the two terms in the expression separately and then find the value of the entire expression $ {(1 + i)^6} + {(1 - i)^3} $ .
Consider $ {(1 + i)^6} = {\left( {{{(1 + i)}^2}} \right)^3} $
Now we will use the algebraic identity $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ to solve the expression. On comparing we can get $ a = 1 $ and $ b = i $ . So, the expression would become $ {(1 + i)^6} = {\left( {{1^2} + {i^2} + 2.1.i} \right)^3} $
$ \Rightarrow {(1 + i)^6} = {\left( {1 + ( - 1) + 2i} \right)^3} $ [Using the value of $ {i^2} = - 1 $ ]
$ \Rightarrow {(1 + i)^6} = {(2i)^3} = 8{i^3} $
$ \Rightarrow {(1 + i)^6} = 8.( - i) = - 8i $
Now consider $ {(1 - i)^3} $ . We can use the algebraic identity for the cube of difference of two numbers $ {(a - b)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3} $ to solve the expression. On comparing the expression with the identity, we get $ a $ as $ 1 $ and $ b $ as $ i $ . Now using the identity, we can write $ {(1 - i)^3} = {1^3} - 3.{(1)^2}.i + 3.(1).{i^2} - {i^3} $ .
On further simplification we get $ {(1 - i)^3} = 1 - 3i + 3{i^2} - {i^3} $ .
$ \Rightarrow {(1 - i)^3} = 1 - 3i + 3.(1) - ( - i) $ [Using the values $ {i^2} = - 1 $ and $ {i^3} = - i $ ]
$ \Rightarrow {(1 - i)^3} = 1 - 3i + 3 + i $
$ \Rightarrow {(1 - i)^3} = 4 - 2i $
So now using the values of these two terms we can find the value of the expression.
On substitution we get $ {(1 + i)^6} + {(1 - i)^3} = - 8i + 4 - 2i $ .
$ \Rightarrow {(1 + i)^6} + {(1 - i)^3} = - 10i + 4 $
Hence the value of the expression $ - 10i + 4 $ .
So, the correct answer is “ $ - 10i + 4 $ ”.
Note: Now, to solve this type of questions we need to know some basic things about $ i $ as it is used to represent an imaginary part of a complex number in the form $ a + ib $ . Moreover, we must know the value of $ i $ as $ \sqrt { - 1} $
Complete step-by-step answer:
The given expression involves the iota $ i = \sqrt { - 1} $ . From its definition we can see that the consecutive powers of iota are $ {i^2} = - 1 $ , $ {i^3} = - 1.i = - i $ , $ {i^4} = {( - 1)^2} = 1 $ . Also note that $ {i^5} = {i^4}.i = 1.i = i $ . So, the values of the next consecutive four powers repeat again. In general, $ {i^{4n}} = 1,{i^{4n + 1}} = i,{i^{4n + 2}} = - 1 $ and $ {i^{4n + 3}} = - i $ . We will use this observation while solving the given expression.
We can use algebraic identities to solve the given expression. We will evaluate the two terms in the expression separately and then find the value of the entire expression $ {(1 + i)^6} + {(1 - i)^3} $ .
Consider $ {(1 + i)^6} = {\left( {{{(1 + i)}^2}} \right)^3} $
Now we will use the algebraic identity $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ to solve the expression. On comparing we can get $ a = 1 $ and $ b = i $ . So, the expression would become $ {(1 + i)^6} = {\left( {{1^2} + {i^2} + 2.1.i} \right)^3} $
$ \Rightarrow {(1 + i)^6} = {\left( {1 + ( - 1) + 2i} \right)^3} $ [Using the value of $ {i^2} = - 1 $ ]
$ \Rightarrow {(1 + i)^6} = {(2i)^3} = 8{i^3} $
$ \Rightarrow {(1 + i)^6} = 8.( - i) = - 8i $
Now consider $ {(1 - i)^3} $ . We can use the algebraic identity for the cube of difference of two numbers $ {(a - b)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3} $ to solve the expression. On comparing the expression with the identity, we get $ a $ as $ 1 $ and $ b $ as $ i $ . Now using the identity, we can write $ {(1 - i)^3} = {1^3} - 3.{(1)^2}.i + 3.(1).{i^2} - {i^3} $ .
On further simplification we get $ {(1 - i)^3} = 1 - 3i + 3{i^2} - {i^3} $ .
$ \Rightarrow {(1 - i)^3} = 1 - 3i + 3.(1) - ( - i) $ [Using the values $ {i^2} = - 1 $ and $ {i^3} = - i $ ]
$ \Rightarrow {(1 - i)^3} = 1 - 3i + 3 + i $
$ \Rightarrow {(1 - i)^3} = 4 - 2i $
So now using the values of these two terms we can find the value of the expression.
On substitution we get $ {(1 + i)^6} + {(1 - i)^3} = - 8i + 4 - 2i $ .
$ \Rightarrow {(1 + i)^6} + {(1 - i)^3} = - 10i + 4 $
Hence the value of the expression $ - 10i + 4 $ .
So, the correct answer is “ $ - 10i + 4 $ ”.
Note: Now, to solve this type of questions we need to know some basic things about $ i $ as it is used to represent an imaginary part of a complex number in the form $ a + ib $ . Moreover, we must know the value of $ i $ as $ \sqrt { - 1} $
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

