Simplify each power of i. i²⁷

Question

Accepted Answer

Hello. Everyone in this video we're gonna be looking at this practice problem where we want to simplify I to the 215th power. So in this case we're working with the imaginary or complex number I and recall that I is equal to the square root of -1. So because I want to simplify it and it's raised to a very high power. Let's think about what happens when I do I squared so I squared is equal to the square root of negative one times the square root of negative one which is equal to negative one. So you can see that when I get squared it actually gives back a real number of negative one. So we're going to use this to help us simplify our expression. We're going to try to pull out as many I squared as we can from this power in order to pull out negative one. So the first thing that we're going to want to do is split up to because to 15 is a odd number. It's not going to give us a even divide if we just pull out or divided by two. So I'm going to redefine I to the 215 Power as I to the 214th Power plus one which is equal to I to the two to the power of 2 14 times I to the first power And now I'm going to work with I two the 14 and try to pull out as many I squares as I can. So I want to pull out as many ice squares as I can from The 14th power, Which means that if I divide to 14 x two, I get 107, so two times 107 is 2 14. So these are the same, these are seeing the same thing just in different formats. And By writing it this way it becomes easier to simplify because we know that I squared is equal to negative one. I'm going to say negative one to the power of 107 and now I have negative one to an odd power. And because I'm raising negative one to an odd power, I will be getting back negative one back because recall it if I raised negative one to an even power, say negative one times negative one which is equal to negative one squared I get back a positive one. However, if I raise negative one To an odd power, say three, I get back a negative number. And that happens because a negative times a negative makes a positive but when you add a third negative we will get back a negative. So that same principle applies. And because 107 is odd, we know that we're going to get back negative one. So we can simplify I to the to 14th power to be negative one and I can plug that back into my expression here. So this becomes negative one times I Which I can just write as negative I so that means the I to the 215th power simplifies to negative I. So that becomes my final simplification for this problem. And you can see that that aligns with answer choice A So hopefully this video helped and see you next time.

Simplify each power of i. i²⁷

Key Concepts

Imaginary Unit (i)

Powers of i and Their Cyclic Pattern

Modular Arithmetic for Exponent Simplification

Watch next

Simplify each power of i. i27

Key Concepts

Imaginary Unit (i)

Powers of i and Their Cyclic Pattern

Modular Arithmetic for Exponent Simplification

Watch next

Simplify each power of i. i²⁷