Question

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

Answer 1

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 $n$