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Question: Evaluate: \( {(1 + i)^6} + {(1 - i)^3} \)...

Evaluate: (1+i)6+(1i)3{(1 + i)^6} + {(1 - i)^3}

Explanation

Solution

Hint : We will first expand the given expression using algebraic identities. Note that i2{i^2} is -1 and i3{i^3} is -1. Also i4{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=1i = \sqrt { - 1} . From its definition we can see that the consecutive powers of iota are i2=1{i^2} = - 1 , i3=1.i=i{i^3} = - 1.i = - i , i4=(1)2=1{i^4} = {( - 1)^2} = 1 . Also note that i5=i4.i=1.i=i{i^5} = {i^4}.i = 1.i = i . So, the values of the next consecutive four powers repeat again. In general, i4n=1,i4n+1=i,i4n+2=1{i^{4n}} = 1,{i^{4n + 1}} = i,{i^{4n + 2}} = - 1 and i4n+3=i{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+(1i)3{(1 + i)^6} + {(1 - i)^3} .
Consider (1+i)6=((1+i)2)3{(1 + i)^6} = {\left( {{{(1 + i)}^2}} \right)^3}
Now we will use the algebraic identity (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} to solve the expression. On comparing we can get a=1a = 1 and b=ib = i . So, the expression would become (1+i)6=(12+i2+2.1.i)3{(1 + i)^6} = {\left( {{1^2} + {i^2} + 2.1.i} \right)^3}
(1+i)6=(1+(1)+2i)3\Rightarrow {(1 + i)^6} = {\left( {1 + ( - 1) + 2i} \right)^3} [Using the value of i2=1{i^2} = - 1 ]
(1+i)6=(2i)3=8i3\Rightarrow {(1 + i)^6} = {(2i)^3} = 8{i^3}
(1+i)6=8.(i)=8i\Rightarrow {(1 + i)^6} = 8.( - i) = - 8i
Now consider (1i)3{(1 - i)^3} . We can use the algebraic identity for the cube of difference of two numbers (ab)3=a33a2b+3ab2b3{(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 aa as 11 and bb as ii . Now using the identity, we can write (1i)3=133.(1)2.i+3.(1).i2i3{(1 - i)^3} = {1^3} - 3.{(1)^2}.i + 3.(1).{i^2} - {i^3} .
On further simplification we get (1i)3=13i+3i2i3{(1 - i)^3} = 1 - 3i + 3{i^2} - {i^3} .
(1i)3=13i+3.(1)(i)\Rightarrow {(1 - i)^3} = 1 - 3i + 3.(1) - ( - i) [Using the values i2=1{i^2} = - 1 and i3=i{i^3} = - i ]
(1i)3=13i+3+i\Rightarrow {(1 - i)^3} = 1 - 3i + 3 + i
(1i)3=42i\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+(1i)3=8i+42i{(1 + i)^6} + {(1 - i)^3} = - 8i + 4 - 2i .
(1+i)6+(1i)3=10i+4\Rightarrow {(1 + i)^6} + {(1 - i)^3} = - 10i + 4
Hence the value of the expression 10i+4- 10i + 4 .
So, the correct answer is “ 10i+4- 10i + 4 ”.

Note : Now, to solve this type of questions we need to know some basic things about ii as it is used to represent an imaginary part of a complex number in the form a+iba + ib . Moreover, we must know the value of ii as 1\sqrt { - 1}