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Question: The value of \({\left( {{i^{18}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^3}\) is equal to ...

The value of (i18+(1i)25)3{\left( {{i^{18}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^3} is equal to
A. 1+i2\dfrac{{1 + i}}{2}
B. 2+2i2 + 2i
C. 1−i2\dfrac{{1 - i}}{2}
D. 2−2i\sqrt 2 - \sqrt 2 i
E. 2−2i2 - 2i

Explanation

Solution

We will write ii in terms of power of 2 and then substitute i2=−1{i^2} = - 1. Then, apply the formula (a+b)3=a3+b3+3a2b+3ab2{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2} to simplify the above equation. Again substitute the value of i2=−1{i^2} = - 1 to write the final answer.

Complete step by step solution:
We have to find the value of (i18+(1i)25)3{\left( {{i^{18}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^3}
Now, we know that i2=−1{i^2} = - 1
The given expression can be written as ((i2)9+(1i25))3{\left( {{{\left( {{i^2}} \right)}^9} + \left( {\dfrac{1}{{{i^{25}}}}} \right)} \right)^3}
We will expression ii in terms of power of 2
((i2)9+(1i.i24))3=((i2)9+(1i(i2)12))3{\left( {{{\left( {{i^2}} \right)}^9} + \left( {\dfrac{1}{{i.{i^{24}}}}} \right)} \right)^3} = {\left( {{{\left( {{i^2}} \right)}^9} + \left( {\dfrac{1}{{i{{\left( {{i^2}} \right)}^{12}}}}} \right)} \right)^3}
Substitute i2=−1{i^2} = - 1
⇒((−1)9+(1i(−1)12))3 ⇒((−1)+(1i))3  \Rightarrow {\left( {{{\left( { - 1} \right)}^9} + \left( {\dfrac{1}{{i{{\left( { - 1} \right)}^{12}}}}} \right)} \right)^3} \\\ \Rightarrow {\left( {\left( { - 1} \right) + \left( {\dfrac{1}{i}} \right)} \right)^3} \\\
We can write 1i=−i\dfrac{1}{i} = - i
⇒(−1−i)3 ⇒(−1)3(1+i)3  \Rightarrow {\left( { - 1 - i} \right)^3} \\\ \Rightarrow {\left( { - 1} \right)^3}{\left( {1 + i} \right)^3} \\\
We know that (a+b)3=a3+b3+3a2b+3ab2{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}
Then, we can simplify the above expression as
⇒(−1)3(13+i3+3(1)2i+3(1)i2) ⇒(−1)3(1+i.i2+3i+3i2)  \Rightarrow {\left( { - 1} \right)^3}\left( {{1^3} + {i^3} + 3{{\left( 1 \right)}^2}i + 3\left( 1 \right){i^2}} \right) \\\ \Rightarrow {\left( { - 1} \right)^3}\left( {1 + i.{i^2} + 3i + 3{i^2}} \right) \\\
Substitute i2=−1{i^2} = - 1 in the above equation.
⇒(−1)3(1−i+3i−3) ⇒(−1)(2i−2) ⇒2−2i  \Rightarrow {\left( { - 1} \right)^3}\left( {1 - i + 3i - 3} \right) \\\ \Rightarrow \left( { - 1} \right)\left( {2i - 2} \right) \\\ \Rightarrow 2 - 2i \\\

Hence, option E is correct.

Note:
A complex number is written in the form of a+iba + ib, where ii is an imaginary number such that i2=−1{i^2} = - 1. Also, we do not write ii in the denominator and then rationalise it by multiplying both in numerator and denominator.