Powers of i - Online Tutor, Practice Problems & Exam Prep

Understanding the powers of the imaginary unit i reveals a repeating cycle of four values: i, -1, -i, and 1. To simplify calculations for higher powers, determine if the exponent is divisible by 4. If yes, the result is 1. If not, find the remainder when divided by 4 to identify the corresponding value in the cycle. For example, i²² yields -1, while i⁶⁷ results in -i. This method streamlines evaluating complex powers of i.

Powers of i

Hey, everyone. We now know that i, our imaginary unit, is equal to the square root of negative one, which is great, but many problems are actually going to take i and raise it to a power. So i may be raised to the second power, the third power, or even something much higher, like i to the 100th power. Now I know that looks a little scary to calculate now, but I'm going to walk you through some lower powers of i that will then allow us to calculate these much higher powers of i super easily. So let's get started. Now, just like we are able to use all of our radical rules for the square roots of negative numbers, we're able to use all of our properties of exponents with our powers of i. So let's go ahead and look at our first power of i, i to the first power.

So if I raise anything to the power of 1, it's just going to be itself, and it's no different here. So i to the first power is just i. Now if I take i squared, if I take i and I square it, I know that i is just the square root of negative one. So if I take that and I square it, I know that squaring a square root just cancels it, and I'm going to be left with negative one. i to the third power, I can go ahead and use some other properties of exponents here and simply expand this into being i squared times i to the first. Now we just calculated both of those powers, so I can just take what I already know those are and plug them in. So i squared, we know, is negative one, and then i to the first power is just i. So I get negative one times i, which is just negative i.

Now i to the 4th power, I can just take i squared and multiply it by i squared because that is equivalent to i to the 4th power. Now I know that i squared is just negative one. So this is just negative one times negative one, which gives me positive one. So that's i to the first is just i, i to the second is negative one, and so on. Let's go ahead and move on to i to the 5th power. Now i to the 5th power, I can expand this out into being i to the 4th times i to the first power. i to the 4th power is a great thing to simplify these down into because since it's just one, it's going to make these really easy. So i to the 4th power again is just 1, and then i to the first is again i. I'm left with 1 times i which gives me i.

Now you might notice that i to the 5th power is the exact same thing we got for i to the first power. So let's see what we get for i to the 6th. So i to the 6th, I can expand that into i to the 4th times i squared. So i to the 4th, again, is just 1, and then i squared is a negative one. 1 times negative one gives me negative one. And this negative one that I got for i to the 6th power is the same thing that I got for i to the second power. So you might notice a pattern here. What do you think i to the 7th will be? Well, it'll actually just be negative i, the exact same thing that i to the third power was. So i to the 8th power is, you guessed it, just one, the exact same thing that i to the 4th power was. So this pattern is actually just going to repeat itself over and over again here. So if I were to take i to the 9th power, that would just restart the pattern. I would get i, i to the 10th power would just be negative one and so on. So that means that every single power of i, no matter how high it is, a 100, a 1000, a million, can always be simplified to one of these four values: i, negative 1, negative i, or 1. That's all for this one. I'll see you in the next one.

Higher Powers of i

Hey, everyone. In the previous video, we learned how to calculate powers of I, and we saw the powers of I just cycle through 4 possible values: I, negative one, negative I, and 1. So we know that any of our powers of I are always going to be one of these four values. But if I'm given something like I to the power of 100, I don't really want to have to count through that cycle until I get to the I to the 100th power. So I'm actually going to show you a much quicker way to evaluate these higher powers of I. Let's go ahead and get started.

So we can actually express all powers of I in terms of I to the 4th power. Now this is going to be really useful for us because I to the fourth power, we know, is just 1. So if I'm given something to the 4th power and multiplying it by itself 5 times. Now since I to the 4th power is just one, this is the same thing as 1 times 1 times 1 times 1 times 1, which we know is just 1. So I to the 20th power is really just 1. But what if I'm given something like I to the power of 22? How might I express that in terms of I to the 4th? Well, I can still start with my 5 powers or my 5 terms of I to the 4th power, and that gets me all the way up to I to the 20th, but I still have 2 left over. So I have all of these I to the 4th terms, which I know just give me 1, but then I have this I squared term. So this is really just 1 times I squared. Now, from our previous video, we know that I squared is just negative one. So this is really just 1 times negative one. Now, 1 times negative one is just negative one. So I22=-1.

But still, if I'm given I to the 100th power, I still don't really want to have to write out I to the 4th power as many times as it takes me to get there. So I'm going to show you an even quicker shortcut of calculating higher powers of I. So to evaluate I raised to a very high power, we're going to ask ourselves one question, and that is, is our power evenly divisible by 4? Now if my answer to that question is yes, then I'm in luck because I'm done and my answer is just 1. So if I'm given something like I to the power of 100 and I ask myself, is this power evenly divisible by 4, my answer is yes because 100 evenly divides by 4 25 times. So I know that my answer is just 1. I100=1.

But if my answer of dividing by 4. So 4 is still super important here, and we're going to use it to find our answer by dividing our power by 4. So let's take a look at that in action. Now, if I'm given I to the power of 22 and I ask myself, is this power evenly divisible by 4? My answer is no. So I need to go ahead and divide this by 4 and find my remainder. Now if I take 22 and I divide it by 4, 4 goes into 22 5 times. 4 times 5 is 20, and I'm left with 2. Now 4 can't go into 2 at all, so 2 is going to be my remainder. This tells me that I22=I2. Now, again, we know that I squared is just negative one. So my answer is negative one. This is the same answer that we got by expanding it in powers of I to the 4th by simply dividing by 4 and looking at our remainder. Let's take a look at another example.

So if I have I to the power of 67 and I ask myself, is this power evenly divisible by 4? My answer is again no. So that means that I need to go ahead and divide by 4 to get my remainder. So if I take 67 and divide it by 4 using long division, 4 goes into 6 one time, and I'm left with 2. If I bring down that 7, 4 goes into 27, 6 times. So 4 times 6 is 24, and that leaves me with 3 left over. And 4 doesn't go into 3 at all, so that means that my remainder is 3. So this I to the power of 67 is really just I to the power of 3. So in this case, my remainder is 3 and we know that I cubed is just negative I. So I67=-I. So my remainder will always be either 1, 2, or 3, so I know that I can easily calculate those values to get my answer. That's all for this one. Let me know if you have questions.

Simplify the power of $i$.

$i^{1003}$

• A. $i$
• B. $-1$
• C. $-i$
• D. 1

Simplify the power of $i$.

$i^{85}$

• A. $i$
• B. -1
• C. $-i$
• D. 1