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Contents

- 1 Imaginary Numbers
- 2 Simplify Imaginary Numbers
- 3 Simplify Imaginary Numbers
- 4 The Product of Negative Radicands
- 5 The Product of Negative Radicands
- 6 Patterns of i
- 7 Patterns of i
- 8 Extending the Pattern of i
- 9 i to the tenth power
- 10 i to the tenth power
- 11 Simplify Powers of i
- 12 Simplify Powers of i Practice
- 13 Simplify Powers of i Practice

When evaluating square roots or other even roots we can't take the root of a negative number. That's because there's no number that can multiply two times, or any even number of times, to have a negative sign. We know a negative times a negative equals a positive number, so any number of two negative signs would always be a positive. So we can't exactly simplify the square root of negative really equals positive 1. And, we also know that the square root of negative 1 does not equal positive 1 since 1 times 1 equals 1. There's no value in the real number system that gives us the square root of negative 1. We'll need something else for this. It turns out that we define it as i, an imaginary number. I allows us to take the square root of negative numbers. For example, if I wanted to take the square root of -4, I would split this radical up, into the square root of -1, and the square root of 4. We know that the square root of 4 is simply 2. And the square of negative 1 really equals i our imaginary number. So, we get the products of 2i. for the second one we have the square root of 3 times the square root of negative 1. This number really represents i. So, we replace the square root of negative 1 with i. We leave the product of the factor in front of the radical. So we'll have I times of the square root of three, our final result. And for our last one, we'll have the square root of negative one times the square root of 12. We can simplify the square root of 12 by breaking it up into the square root of four times the square root of three. We know the square root of negative one is really equal to I. So we write that here. And then we take the square root of 4 which is 2. We multiply these two together to get 2i times the square root of 3. In a lot of ways we treat i just like a variable. We list it after the product of any numerical factors but before the radical sign. Now that we've seen a couple of cases of simplifying negative radicals with i let's see if you have the hang of it.

What do you think each of these radicals would equal? Simplify them using what you know about I, and what you know about simplifying numbers underneath the radical. Good luck.

We'll have 5i, 3i root 2, 3i root 3 and 4i root 3. Great effort if you got at least one of those correct. For the first one we simply take the square root of one we'll split up negative 18 into 9 times 2. times the square root of negative 1. We know the square root of 9 equals 3, and the square root of 2 needs to stay as the square root of 2, since 2 can't be simplified. We've replaced the square root of negative 1, with i, leaving us with 3 times i root finally, for this last radical, we'll break up the square root of negative 48 into 16 times 3 times the square root of negative 1. We know the square root of can't be simplified. We replace the square root of negative 1 with i and then multiply 4 and i to get 4i root 3. Our final answer. There are of course other ways to simplify this last expression this would be the quickest.

We know that i equals the square root of negative 1. This must mean that if we square i then that means we would square the square root of negative 1. We know that this power and this square root undo one another. So we would just be left with negative 1. Knowing that i squared equals negative 1, wouldn't you think the square root of negative 4 times the square root of negative 9 would equal? Choose the best answer out of these.

It turns out that this answer would be negative 6. Let's see why. You might have thought to multiply these together to get the square root of positive 36 which equals 6. But we know that's not true since 6 is a real number and these two are imaginary numbers. The product rule does not hold for imaginary numbers. It only holds for real numbers. This was our product rule, and before X and Y had to be greater than zero, that was so that we would leave these radicals for real numbers rather than real numbers. instead of multiplying these radicans together, let's simplify each one by using our imaginary number I. So we'll have 2 I here and 3 I for here. Multiplying, we will get 6 I squared... We can replace this i squared with its corresponding value negative answer.

I turns out that this i has an interesting pattern to it. We can simply any power of i by following this pattern. First, we need to remember that i equals the square root of negative 1. This means if we square i, we also square the square root of negative 1. So i squared would equal negative 1. To find the value of i cubed, we multiply i squared by i. Since the value of i squared is negative 1, we simply have negative 1 times i, or negative i. So we know i cubed really simplifies to negative i. So let's use a similar approach to find i to the 4th. We'll split it up into two powers of i that we know, i squared and i squared. Since i squared equals negative 1, we have negative 1 twice. Negative 1 times negative 1 equals positive 1. So i to the 4th will just simplify to positive 1. Let's see if you can extend our pattern. What do you think i to the 5th, i to the 6th, i to the 7th, and i to the 8th would be? If you get the square root of negative 1, I want you to just enter i for your answer. And as a hint to get you started, you can rewrite i to the 5th as i to the 4th times i. Think about if you can replace any of these values.

It turns out that i to the 5th equals i, i to the 6th equals negative 1, i to the 7th equals negative i, and i to the 8th equals positive 1. Fantastic pattern solving, if you found those. Now, I know these are pretty tough. So, it's okay if you stumbled. Let's see how to do it. For the first one, we have i to the 4th times i. We know i to the 4th already. That equals 1. So, we can replace i to the 4th with 1. This leaves us with 1 times i, which equals i. For i to the 6th, we can rewrite it as i to the 4th times i squared. We know i to the 4th equals positive 1 from before. And we know that i squared really equals negative 1. So, we can replace that here. We multiply these numbers together to give us negative 1. For i to the 7th, we break it down into i to the 4th times i cubed. i to the 4th equals 1, and we know i cubed is really equal to negative i. So, that's how we simplify i to the 7th to be negative i. And finally, for i to the 8th, we'll have i to the 4th times i to the 4th. Both of these equal 1. So, we'll have 1 times 1, which equals 1.

So, I hope you're starting to see the pattern. We have i and then i squared equals negative 1, i cubed equals negative i, and i to the 4th equals positive negative 1, i to the seventh equals negative i, and i to the eighth equals positive 1. Taking out all the simplification, it's more clear to see the pattern. We have i, negative 1, negative i, positive 1. I, negative 1, negative i, positive 1.

So, based on our pattern, what do you think i to the 10th would be? Would it be i, negative 1, negative i, or positive 1?

We know that for the pattern of i. It repeats after each fourth power. So for i to the tenth, we would go to the cycle twice. And then we would start over. We would have i, then negative 1. I to the tenth is really the same thing as i to the sixth and i squared. They all equal negative 1. We can think of this a lot like the arithemetic on a clock. A clock cycles every 12 hours, the hour hand passes all the numbers, and gets to 12 and then starts over. Well, that's similar to what's happening with i, so we'll start with i and then we'll have i squared I-cube, either the fourth, either the fifth, either the sixth, either the seventh, either the eighth, either the ninth, and finally either the tenth. This is how we know either the tenth is really equal to i squared, which equals negative 1. Our pattern repeats every four patterns of i

We can always simplify powers of i since the pattern continues to repeat itself. To find any power of i we simply divide the power of i by four and look at the remainder. The remainder will either be 0, 1, 2, or 3 and will determine the power of i. For example i to the 12th, we would take the power of 12 and divide it by 4. We know 12 divided by 4 equals 3 with a remainder of 0. This remainder becomes the new power for i. So we know i to the 12th is really equivalent to i to the 0, and anything to the 0 power equals 1. This should also make sense since i to the 12th can be written as i to the 4th, times i to the 4th, times i to the 4th. This means we really have i to the fourth repeated a bunch of times, or really just the multiplication of positive 1. It's easiest to simplify any power of i by dividing the power by 4 and looking at the remainder, this will be our new power of i. Once we have our power of i, we look back to see what it would equal. In this case, we don't need to, since we have i to the 0, and anything raised to the zero power equals 1. We know our pattern repeats every 4. So if we worked backwards with our pattern. We know i to the 0 would equal positive 1. So this should definitely make sense with our pattern. So, if I wanted to simplify i to the 39th, I would do 39 divided by 4. We know this is 9 remainder 3. This is what I'm interested in. The remainder. So I know i to 39th is really equal i to the third power. We're going to cycle to this pattern of four. Nine times and then ended up with i cube. We know i cubed is really equal to negative i. So this power simplifies to negative i. If I want to simplify i to the 50th. I would divide 50 by 4 which is 12 remainder

Now that we've seen how to simplify powers of i, I want you to try and simplify each of these. You want to to use the idea we just covered about remainders and maybe use this table as a reference. Good luck.

Here are the four answers. Great work if you found those. We know what pattern of I repeats every fourth power, so we want to take this power of 21 and divide it by 4. Doing so we get 5 remainder 1. This means I to the 21st power is equivalent to I to the 21, which we know is really just I. For the second one we'll take the power of 11 and divide it by 4... Which is 2 remainder 3. So we know i to the 11th, is equivalent to i to the 3rd, which equals -i. Since i to the 48th is a multiple of 4, we end up on a power of i to the 4th, which is really just 1. When thinking about it with a remainder, we know 48 divided by 4 equals 12 remainder 0. i to the 48th is equal to i to the 0, which equals 1. And finally for i to the 66th, we divide the power by 4 to get 16 remainder 2. This means that we cycle through the patterns of i 16 times and then wind up with i squared, which is equal to negative 1. If this is still giving you trouble try repeating these values all the way up to i to the 11th. You'll start seeing the pattern repeat and hopefully you'll catch the hang of it.