Solveeit Logo

Question

Question: How do you simplify \({i^{37}}\)?...

How do you simplify i37{i^{37}}?

Explanation

Solution

In the question above we have a value of an imaginary number that states that it is with the power, i37{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 - 1.
This means that when multiplied with its fourth power, the number will be equal to 11. 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 i37{i^{37}} and we have to find its value, for that we will start by recalling that,
i=1\Rightarrow i = \sqrt { - 1}
Now, we know that the value of the imaginary number is square root of 1 - 1, using this for further calculation we get,
i2=1\Rightarrow {i^2} = - 1
Therefore, the value further will be,
i3=1×1\Rightarrow {i^3} = - 1 \times \sqrt { - 1}
And finally, the fourth power will hold the value like,
i4=1×1\Rightarrow {i^4} = - 1 \times - 1
This will be of the value,
i4=1\Rightarrow {i^4} = 1
Whenever the exponent on ii is a multiple of 44, it will evaluate to 11, as imaginary numbers follow a pattern.
3636 is a multiple of 44, so we know i36=1{i^{36}} = 1. We can rewrite i37{i^{37}} as,
i36×i\Rightarrow {i^{36}} \times i
This can be substituted as,
1×i\Rightarrow 1 \times i
This will multiply to,
i\Rightarrow i
Thus, i37=i{i^{37}} = i

Therefore, the value of the imaginary number i37=i{i^{37}} = i

Note: An imaginary number is a Complex number that can be written as a real number multiplied by the imaginary unit II, which is defined by its property i2=1{i^2} = - 1. The square of an imaginary number bi is b2 - {b_2}. For example, 5i  5i\; is an imaginary number, and its square is 25 - 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.