How do you simplify ${i^{37}}$?

Hint: In the question above we have a value of an imaginary number that states that it is with the power, ${i^{37}}$, and we have to simplify it. While simplifying it, one must know that for an imaginary number, the value of its square is always $ - 1$.
This means that when multiplied with its fourth power, the number will be equal to $1$. Using this conclusion in mind, we will solve this question, following the steps and showing required proof.

Complete step-by-step solution:
We have an imaginary number ${i^{37}}$ and we have to find its value, for that we will start by recalling that,
$ \Rightarrow i = \sqrt { - 1} $
Now, we know that the value of the imaginary number is square root of $ - 1$, using this for further calculation we get,
$ \Rightarrow {i^2} = - 1$
Therefore, the value further will be,
$ \Rightarrow {i^3} = - 1 \times \sqrt { - 1} $
And finally, the fourth power will hold the value like,
$ \Rightarrow {i^4} = - 1 \times - 1$
This will be of the value,
$ \Rightarrow {i^4} = 1$
Whenever the exponent on $i$ is a multiple of $4$, it will evaluate to $1$, as imaginary numbers follow a pattern.
$36$ is a multiple of $4$, so we know ${i^{36}} = 1$. We can rewrite ${i^{37}}$ as,
$ \Rightarrow {i^{36}} \times i$
This can be substituted as,
\[ \Rightarrow 1 \times i\]
This will multiply to,
\[ \Rightarrow i\]
Thus, \[{i^{37}} = i\]

Therefore, the value of the imaginary number \[{i^{37}} = i\]

Note: An imaginary number is a Complex number that can be written as a real number multiplied by the imaginary unit \[I\], which is defined by its property ${i^2} = - 1$. The square of an imaginary number bi is \[ - {b_2}\]. For example, \[5i\;\] is an imaginary number, and its square is \[ - 25\]. By definition, zero is considered to be both real and imaginary. The set of imaginary numbers is sometimes denoted using the blackboard bold letter.