Question

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

Answer 1

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

Complete step by step solution:
We have to find the value of ${\left( {{i^{18}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^3}$
Now, we know that ${i^2} = - 1$
The given expression can be written as ${\left( {{{\left( {{i^2}} \right)}^9} + \left( {\dfrac{1}{{{i^{25}}}}} \right)} \right)^3}$
We will expression $i$ in terms of power of 2
${\left( {{{\left( {{i^2}} \right)}^9} + \left( {\dfrac{1}{{i.{i^{24}}}}} \right)} \right)^3} = {\left( {{{\left( {{i^2}} \right)}^9} + \left( {\dfrac{1}{{i{{\left( {{i^2}} \right)}^{12}}}}} \right)} \right)^3}$
Substitute ${i^2} = - 1$
$
   \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 $\dfrac{1}{i} = - i$
$
   \Rightarrow {\left( { - 1 - i} \right)^3} \\
   \Rightarrow {\left( { - 1} \right)^3}{\left( {1 + i} \right)^3} \\
$
We know that ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$
Then, we can simplify the above expression as
$
   \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 ${i^2} = - 1$ in the above equation.
$
   \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 $a + ib$, where $i$ is an imaginary number such that ${i^2} = - 1$. Also, we do not write \[i\] in the denominator and then rationalise it by multiplying both in numerator and denominator.