Question

How do you simplify ${i^{18}}$?

Answer 1

Hint: Let us first know what a complex number is. Complex numbers are any number in the form of $Z = a + ib$ where $a$ and $b$ are real numbers and $i$ is an imaginary number known as iota. It is equal to $\sqrt { - 1} $. Complex numbers are denoted as $Z$.
It must be known that when $i$ is raised to certain powers, it assumes some values which are given as:
$i = \sqrt { - 1} $
${i^2} = - 1$ (As ${i^2} = {(\sqrt { - 1} )^2} = - 1$)
${i^3} = - i$ (As ${i^3} = {i^2} \times i = - 1 \times i = - i$)
${i^4} = 1$ (As ${i^4} = {i^2} \times {i^2} = - 1 \times - 1 = 1$)
We will use these $4$ values of $i$ to solve the given question.

Complete step by step solution:
Given expression is ${i^{18}}$. We will now simplify it by breaking the exponent $18$into factors, such that
$ \Rightarrow {i^{18}} = {i^{2 \times 3 \times 3}}$
We will now separate the factors of the exponent into $2$ and all other factors, such that
$ \Rightarrow {i^{18}} = {i^{2 \times 9}}$
We will now use one of the law of exponents which states ${x^{p \times q}} = {({x^p})^q}$, such that
$ \Rightarrow {i^{18}} = {({i^2})^9}$
On substituting the value of ${i^2} = - 1$, we will get
$ \Rightarrow {i^{18}} = {( - 1)^9}$
It must be known that if $ - 1$ is raised to the power of an odd number, we will get $ - 1$ itself as the answer. Since $9$ is an odd number, therefore we will get
$ \Rightarrow {i^{18}} = - 1$

Hence, on simplifying ${i^{18}}$, we get $ - 1$ as the answer.

Note:
The given question can also be solved in an alternate way. We know that ${i^4} = 1$. So we will simplify the exponent $18$of the expression ${i^{18}}$ in the form of $4m + n$ where $m$ and $n$ are whole numbers. Hence on simplifying, we will get
$ \Rightarrow {i^{18}} = {i^{4 \times 4 + 2}}$
We will now use one of the law of exponents which states ${x^{p + q}} = {x^p} \times {x^q}$, such that
$ \Rightarrow {i^{18}} = {i^{4 \times 4}} \times {i^2}$
Again we will use one of the laws of exponents which states ${x^{p \times q}} = {({x^p})^q}$, such that
$ \Rightarrow {i^{18}} = {({i^4})^4} \times {i^2}$
We know that ${i^4} = 1$. Therefore on substituting this value in the above equation, we will get
$ \Rightarrow {i^{18}} = {(1)^4} \times {i^2}$
$ \Rightarrow {i^{18}} = 1 \times {i^2}$
So ${i^{18}}$ gets reduced as given,
$ \Rightarrow {i^{18}} = {i^2}$
Now we know that ${i^2} = - 1$. Therefore on substituting this value in the above equation, we will get
$ \Rightarrow {i^{18}} = - 1$
Hence on simplifying ${i^{18}}$, we got $ - 1$.