Question

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

Answer 1

Hint: We can simplify the expression using the powers of \[{\text{i}}\]. We know that i raised to powers which are multiples of 4 are equal to 1. Using this we can simplify the powers of i to get the required value of the expression.

Complete step by step solution: We have the expression
${\left( {{i^{18}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^3}$
We can write the power of ${\text{i}}$ as multiples of 4 and its remainders. Then we get
\[
  {\left( {{i^{18}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^3} = {\left( {{i^{4 \times 4 + 2}} + {{\left( {\dfrac{1}{i}} \right)}^{4 \times 6 + 1}}} \right)^3} \\
   = {\left( {{{\left( {{i^4}} \right)}^4} \times {i^2} + \dfrac{1}{{{{\left( {{i^4}} \right)}^6} \times {i^1}}}} \right)^3} \\
 \]
We know that ${i^4} = 1$ and ${i^2} = - 1$. By using this relation, we get,
\[
   = {\left( {{{\left( 1 \right)}^4} \times - 1 + \dfrac{1}{{{{\left( 1 \right)}^6} \times {\text{ }}i}}} \right)^3} \\
   = {\left( { - 1 + \dfrac{1}{i}} \right)^3} \\
 \]
We know that \[\dfrac{1}{i} = - i\]
\[ = {\left( { - 1 - i} \right)^3}\]
We can take the negative sign outside. We get,
\[ = - {\left( {1 + i} \right)^3}\]
Using cubic expansion ${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$, we get,
\[ = - \left( {{1^3} + 3 \times {1^2} \times i + 3 \times 1 \times {i^2} + {i^3}} \right)\]
Using the relations ${i^3} = - i$ and ${i^2} = - 1$, we get,
\[
   = - \left( {1 + 3i - 3 - i} \right) \\
   = - \left( { - 2 + 2i} \right) \\
   = 2 - 2i \\
 \]
Therefore, the value of the expression is equal to $2 - 2i$

So, the correct answer is option E.

Note: In this problem, we are only using the concept of powers of ${\text{i}}$ and exponents. We use the concept of the division algorithm to change the power of i to multiples of 4 and its remainder. The basic idea of division algorithm is that every number n can be written in the form ${\text{n = mq + r}}$, where r is remainder, q is the quotient, and m is the divisor.