How do you simplify $ {i^{999}}? $

Hint: We know that the above given question is in exponential form. An exponent refers to the number of times a number is multiplied by itself. There is base and exponent or power in this type of equation. Here, in the given question $ (i) $ is the base and the number $ 999 $ is the exponential power. Here in this question $ i $ is an imaginary number part of the complex number. It is also called iota. The value of $ i $ is $ \sqrt { - 1} $ .

Complete step by step solution:
As we know that the value of $ i $ is $ \sqrt { - 1} $ , so if we square it the value will be $ {i^2} = - 1 $ .
To get the positive value of iota we have to square it again i.e. $ {i^4} = 1 $ . As we can see that the value is positive, so to make the solution easier we will take the remainder of the power divided by $ 4 $ , instead of using the actual power.
So the power when divided is $ \dfrac{{999}}{4} = 249 $ and the remainder is $ 3 $ . By using the above application we can write the question as $ {({i^4})^{249}} \times {i^3} $ .
By putting the values we have: $ {(1)^{249}} \times {i^3} $ .And $ {i^3} $ can be written as $ - i. $
Hence the required answer of the exponential form is $ - i. $
So, the correct answer is “ $ - i. $ ”.

Note: In the above solution we have used another exponential formula. As we know that as per the property of exponent rule if there is $ {(ab)^m} $ then it can be written as $ {a^m} \times {b^m} $ . Also we should note that the powers of $ i $ have a repetitive cyclic nature. The formula applied before is true for all real values of $ m $ and $ n $