WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to take a look at complex number.
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I held off on this very special type of number so you could build a lot of things about them that you will see in the section
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such as multiplying them together which will look like multiplying polynomials and rationalizing denominator will look a lot like the division process.
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Here are some other things that we will cover.
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First we will get into a little bit about that the vocabulary from the complex numbers.
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You have heard about imaginary numbers and you will see how that works with complex numbers.
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We will learn about the real and the imaginary part of complex numbers.
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Many of the properties involving these complex numbers are the same properties that you have seen before for real numbers.
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Then we will get into the nuts and bolts on how you can start combining these things.
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That means adding, subtracting, multiplying and dividing our complex numbers.
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Near the very end we will also see some nice handy ways that we can simplify powers of (i)
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so you can always bring them down to the most simplest form.
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An imaginary number (i) is defined as the √-1.
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It seems like a small definition and it does not seem to be very useful, especially if we want talk about lots of different numbers.
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By defining it in this way, the √-1 it can actually express the root of any negative number using this imaginary number (i).
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Here is a quick example to show how it works.
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Suppose I’m looking at the √-7 using some of our properties we can go ahead and break it up into it -1 × 7.
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And then break up the root over each of those parts.
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You will see the definition shows up right there.
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I'm taking the √ -1.
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It is that piece that gets turned into an (i) and now I'm representing this number as (i) × √7.
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Think of doing this for many other numbers where you have a negative underneath the square root.
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If I had -23 underneath the square root that will be (i)√23 and a negative comes out as the (i).
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√ -16 would be (i) × 4 or just 4i.
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How those imaginary numbers relate to our complex numbers?
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Complex number is a number of the form A + B(i).
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You can see that the imaginary number is sitting right there, right next to that B.
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Those other values A and B are both real numbers.
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We call A the real part of the complex number and B the imaginary part, since it is right next to the imaginary number.
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You can represent a lot of different numbers using this nice complex form.
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Maybe I will have a number like 3 – 5i that would be in a nice complex form.
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I can read off the real and imaginary part.
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We could throw in some fractions and in here as well, so maybe ½ × 2/3(i).
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We can also have complex numbers like 0 + 3(i).
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We would probably would not leave it like that, something like this would probably we will write as 3(i).
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Notice how we can still say it is in a complex form where the real part is 0.
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The good part about these complex numbers is they will obey most of the usual laws that you are familiar with.
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Our commutative, associative, and distribution, all of those will work with our imaginary numbers as well.
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The things that we could do with real numbers, we can do with imaginary numbers as well.
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Some of the things that we will have to watch out for is when we get down in the simplification process, so keep a close eye on that.
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Let us get into how you can start combining these numbers together.
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To add or subtract complex numbers this is a lot like having like terms.
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Be very careful to watch the signs when combining these numbers together.
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To show you how this works I have 2 – 5(i) that complex number + 1 + 6(i).
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Since I'm adding I'm going to go ahead and drop my parentheses here and just figure out which terms I should add together.
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Notice how all of our real parts I can go ahead and put those together and I can go ahead and collect together the imaginary parts.
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That is what I'm talking about when I say adding like terms.
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The real ones 2 + 1 would be 3 and the imaginary I have -5(i) -6(i) that will give me -11(i).
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I will write those as 3 -11(i).
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When you are working with subtraction we have to be very careful with your signs.
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Let me show you, suppose I have 3 + 2(i) and now I'm subtracting a second number 4 + 7(i).
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Parentheses will come in handy because I need to distribute this negative sign to both parts of that second complex number.
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I will look at this as 3 + 2(i) – 4 – 7(i) and now I can see the parts and the like terms that I need to go ahead and put together.
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3 – 4 = -1 and 2 – 7(i) = -5(i)
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This would be written as -1 – 5(i)
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Again, combine those like terms.
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To multiply complex numbers think of multiplying together two binomials.
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It looks a lot like the same process.
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When we are multiplying together two binomials we have great way of remembering that we can use the method of foil to accomplish that.
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There is one additional thing you have to remember, when you have (i)², we can go ahead and reduce that to -1,
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that seems a little odd but let us see why that works.
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If I'm looking at the √-1 and that is our (i) and suppose I take that and I square it.
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I want to square that square root I will get -1 on the left.
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That reveals that (i)² = -1.
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Watch for that to show up in the simplification process.
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Let us go ahead and foil out these two complex numbers.
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2 – 5(i) multiplied by 1 – 6(i)
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We will begin by multiplying the first terms 2 × 1 = 2.
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We will move on to those outside terms 2 × -6(i) = -12(i).
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On the inside terms -5(i) and now let us go ahead and do the last terms -5(i) × -6(i),
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minus × minus would be +, 5 × 6 is 30 and now there is my (i)².
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We will go along and we will start crunching things down like we normally would do if they were a couple binomials.
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2 – 17(i) now comes this interesting part this (i)² here is the same as -1.
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We will go ahead and rewrite it as a -1.
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You can see there is more than we can do to simplify this.
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I’m not only multiplying the 30 × -1, but then I can go ahead and combine it with the two out front.
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2 – 17(i) - 30 and then let us combine these guys that will give us -28 – 17(i).
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This one is good.
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With many of these problems where combining complex numbers you know that you are done
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when we finally get into that nice complex form.
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You can easily see the real and imaginary parts.
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To go ahead and divide complex numbers, we want to multiply the top and the bottom by the complex conjugate.
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The complex conjugate of a number will look pretty much the same but it will be different in sign.
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If I'm looking at the complex conjugate of A + B(i) that will be –B(i).
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Let us just do some quick example.
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Let us see if I have 3 + 2(i) its complex conjugate would be 3 – 2(i).
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If I had -7 – 3(i), the complex conjugate would be -3 + 3(i).
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The only thing that is changing is that negative sign or that sign on the imaginary part.
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This will have the feel of rationalizing the denominator that is why we had to cover a lot of that information before working on these complex numbers.
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I’m going to look at the bottom of this particular division problem 4 + 2(i) ÷ 3 + 5(i).
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I'm going to find the complex conjugate of 3 + 5(i), that would be 3 – 5(i).
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Once we find that complex conjugate we will multiply on the top and on the bottom of our complex fraction.
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Multiply on top, multiply on the bottom.
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This one involves quite a bit of work.
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I have two terms in my complex number, I will have to foil the top and we will also have to foil out the bottom.
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This will make things look a lot more complicated at first but if you just go through the problem carefully you should do fine.
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Let us start with the top.
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Taking our first terms I have 4 × 3 and that would give me 12.
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The outside terms would be 4 × -5 (i) = -20(i).
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On the inside 2(i) + 3(i) then would be 6(i) and our last terms 2 × -5 = -10(i)².
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Do not forget to multiply those (i) together as well.
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On the button 3 × 3 = 9, outside will be -15(i), inside would be 15(i).
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And then we have 5(i) × -5(i) = -25(i)².
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Some things you want to notice after you go through that forming process.
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We are using that complex conjugate on the bottom, the outside and inside terms will end up canceling.
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If they do not cancel, make sure you have chosen the correct complex conjugate.
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Notice that we have a couple of (i)² showing up on the top and bottom.
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That comes from the (i) being multiplied together.
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Each of these will be need to change into -1.
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Let us go through and do those two things and clean this problem a little bit.
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I have 12 – 20 + 6(i) = -14(i), -10 × -1 ÷ 9 my outside and inside terms on the bottom cancel -25 × -1.
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We can see the beauty of using that complex conjugate.
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On the bottom we no longer have any more imaginary numbers.
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That one I was talking about how I have the feel of rationalizing the denominator and we are getting rid of those roots in the bottom.
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Here we got rid of the imaginary numbers in the bottom.
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Let us continue to simplify and see where we can take this problem.
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12 – 14(i) + 10 ÷ (9 + 25)
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12 + 10 = 22 – 14(i) ÷ 34
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It is tempting to try and stop right there but do not do it, we want to continue doing this entire problem
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and try to get it in that nice complex form where I can see the real and imaginary part.
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To get in that form I’m going to use the part where I breakup the 34 and 22 and under the 14.
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It looks a lot like when we are taking polynomials when we are dividing by a monomial.
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This will be 22 ÷ 34 -14(i) ÷ 34.
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This particular one, both of those fractions can be reduced.
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I will divide the top and the bottom by 2, 11/17 – 7(i) ÷ 17.
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This one is actually completely done.
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I can easily see my real part over here and my imaginary part -7/17.
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That is I know that this one is done.
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Division is quite a tricky process remember to multiply by the complex conjugate of the bottom
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and then work very carefully to simplify the problem from there.
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It should be in its complex form when you are all done so, you can see the real and imaginary parts.
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After working with doing some combining things with these various complex imaginary numbers,
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you may have noticed that you rarely see any higher powers of (i).
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In fact the largest part of (i) that we came across was i² and we turned it immediately into a -1.
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The interesting part is you can usually take much higher powers of (i) and end up simplifying them down.
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To see out why this works for some much higher powers, let us just take a bunch and start picking them apart.
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We see many instances while we are working with (i) and (i) was just equal √1 which of course we call (i).
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When we have i² and then we thought about squaring the √-1, so we got -1 as its most simplified result.
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We will see what will happen with i³.
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One way that you could interpret this would be i² × (i).
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The reason for doing that is because you can then take the i² sitting right here and simplify that.
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That will give us -1 × (i) the final result of that would be –i as it is simplest form.
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Let us choose that same idea and see if we can go ahead and simplify i⁴.
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That will be i² × i².
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Both of the i² = -1
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I have -1 × -1 = 1
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Take note how all of these are simplifying to something else that does not have any more powers on them.
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Let us do a few more that hopefully we can develop a much easier or a shortcut way of doing this process.
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Let us go on to i⁵.
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This can be thought of as i⁴ × (i).
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We have already done i⁴, it is way back here, that is equal to just 1.
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1 × (i) I know this simplifies to (i).
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On the next one i⁶, i⁴ × i².
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We have already done i⁴ that is just 1.
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I will put in i², I will just write across from it -1 so 1 × -1 = -1, very interesting.
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I think we are starting to see a pattern of some of these (i).
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Notice how they are looking exactly the same as the ones on the other side here.
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Let us see if that continues.
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I⁷ will be i⁴ × i³, 1 × i³ = -i.
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Maybe the last one, i⁴ × i⁴, both of those are equal to one, this is equal to 1 as well.
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Notice how we did starting to see same numbers.
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There are some good patterns we can see, I mean we got past i⁴ and all of them contained an i⁴
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so we are getting this similar pieces right here.
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If I was even to continue on to i⁹ then I will have even more groups of i⁴ and I could see that it would simplified down.
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All of those much higher powers will end up simplifying down to something which does not have any more powers whatsoever.
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We can take as much farther and develop a nice shortcut formula so we do not always have to string out into a bunch (i).
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Let us see how to do that.
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Looking at all of these examples you can see that there is a pattern on how they simplify.
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In fact they repeats in blocks of 4 and you start with (i) then it goes -1, then –I, then 1, it just repeats over and over again.
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To figure out where in that pattern you are at, you can take the exponent and simply divide it by 4 and observe what the remainder is.
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From the remainder you will know exactly what it will simplify to.
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I have made a handy table to figure out what that looks like
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If I’m looking at (i) to some large power, I will take the power and divide it by 4
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and the remainder happens to be 3, according to my chart here it will simplify to –i.
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Let us do a quick example to see how this works and why it works.
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Let us take i^13.
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If I wanted to I could imagine a whole bunch of i⁴ in here.
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i⁴, i⁴, i⁴ and then i¹.
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If we add up all those exponents they add up to 13.
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You can see that most of them are gone since i⁴ = 1.
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1 × 1 × 1 × (i) =(i)
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Let us use our shortcut formula to figure out the same thing.
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In the shortcut we take the exponent and divide that by 4.
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13 ÷ 4, 4 goes in the 13 exactly 3 times.
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Notice why that is important, that is exactly how many bunches of i⁴ I have.
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When you are going through that division by 4 process you are counting up all of the i⁴ that will simply break down and simplify into 1.
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It goes in there 3 times, 3 × 4 = 12 and I will subtract that away getting my remainder of 1.
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That remainder tells me how many extra (i) I still have left over.
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From there I can see I have only one (i) and I know it simplifies down to (i).
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I get the same exact answer as the 4, i^13 = (i).
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Now that we know a lot more about these complex numbers let us go through some various examples
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of putting them together and see how we can simplify powers of (i).
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This first one we want to do a subtraction problem 4 + 5(i) – 6 – 3(i).
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With the addition and subtraction we are looking to add like terms.
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Subtraction is a tricky one, you have to remember to distribute our sign onto both parts of that second complex number.
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I’m looking at 4 + 5(i) -6 + 3(i)
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I can see my like terms, let us put together 4 and -6, and 5 and 3.
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4 - 6 = -2, 5(i) + 3(i) = 8(i)
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We can see that it is in that final complex form and we know that this one is done.
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In this next example, we want to multiply two complex numbers.
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Think foil.
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Be on the watch out for additional things that we will need to simplify such as the i².
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Okay, starting off, multiplying our first terms together we will get 1.
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Our outside terms we will multiply that will be 3(i) then we can take our inside terms 2(i) and then finally take our last terms 2(i) × 3(i) + 6i².
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We go through and start combining everything we can.
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1 + I have my like terms here, so 3 + 2 = 5(i) + 6 and take the i² and turn it into -1.
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Let us continue 1 + 5(i) - 6 now we can see we have just a couple of more things that we can go ahead and combine.
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This will be -5 + 5(i) and now I can easily see my real and imaginary part, so I know that one is done.
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Let us divide these two complex numbers.
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I have 2 – 5(i) ÷ 1 – 6(i)
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This is the lengthy process where we find the complex conjugate of the bottom of if we multiply the top and bottom of our fraction by that.
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Let us first find that complex conjugate.
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That will be 1 + 6(i).
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We are going to use that on the top and on the bottom.
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We have to remember that since we are multiplying we will have to foil out the top and the bottom.
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Foil top and foil bottom.
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Let us see what the result of that one looks like.
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I’m looking at the top, first terms 2 × 1 =2, outside terms would give me 12(i), inside terms -5(i), and last terms -30(i)².
00:24:39.500 --> 00:24:43.100
Lots of terms in there, be very careful and keep track of them all.
00:24:43.100 --> 00:24:56.000
On the bottom, first terms 1, outside 6i, inside -6i, and our last terms -36i².
00:24:56.000 --> 00:24:58.300
It is looking much better, looking pretty good.
00:24:58.300 --> 00:25:02.100
If we do this correctly, we should not have any more powers of (i) in the bottom.
00:25:02.100 --> 00:25:05.600
Let us see what else we can simplify.
00:25:05.600 --> 00:25:10.600
We will go ahead and combine these 2i on the top and we will combine these 2i of the bottom
00:25:10.600 --> 00:25:17.800
and then we will put these i² to make them both -1.
00:25:17.800 --> 00:25:27.200
2 and then we will have 7(i) - 30 × -1.
00:25:27.200 --> 00:25:33.100
On the bottom 1 – 36 -1.
00:25:33.100 --> 00:25:37.800
It looks like we have indeed accomplished our goal there is no longer imaginary numbers on the bottom.
00:25:37.800 --> 00:25:53.000
We continue putting things together 2 + 7(i) + 30 since the negative × negative is a positive and 1 + 36.
00:25:53.000 --> 00:26:01.800
Just a few more things to put together 32 + 7(i) ÷ 37.
00:26:01.800 --> 00:26:10.000
This point it is tempting to stop it but keep going until you get it into its nice complex form, we can see that real and imaginary part.
00:26:10.000 --> 00:26:19.400
I’m going to give 37^32 and 37⁷.
00:26:19.400 --> 00:26:26.700
7 /37(i) this one looks much better.
00:26:26.700 --> 00:26:34.900
Sometimes you can go ahead and reduce those fractions but this one the way it is.
00:26:34.900 --> 00:26:41.900
In this last problem we will go ahead and see if we can simplify some very large powers of (i).
00:26:41.900 --> 00:26:47.600
Since they are very large powers we will go ahead and use our shortcut to take care of that process.
00:26:47.600 --> 00:26:51.600
In the first one I have i^107.
00:26:51.600 --> 00:26:58.600
Let us start off by taking that exponent, 107 and we will divide it by 4.
00:26:58.600 --> 00:27:03.300
This will figure out all the bunches of i⁴ that we have hiding in there.
00:27:03.300 --> 00:27:09.400
4 of those in the 10, twice and we get 8.
00:27:09.400 --> 00:27:14.800
Subtract them away I will end up with 27.
00:27:14.800 --> 00:27:24.700
That goes in there 6 times and 6 × 4 = 24 and then we can subtract that away.
00:27:24.700 --> 00:27:32.900
I'm down to 3 and that is smaller than 4 so I know that 3 is my remainder.
00:27:32.900 --> 00:27:40.100
Think back, if my remainder is 3 then what does this entire thing simplify to?
00:27:40.100 --> 00:27:44.400
It simplifies to –i.
00:27:44.400 --> 00:27:54.400
If you ever forget the way you use that table, you can always build it very quickly by doing just the first few values of (i).
00:27:54.400 --> 00:28:01.000
(i) = (i), i² = -1, and we have –I and 1.
00:28:01.000 --> 00:28:05.900
You quickly have built your table.
00:28:05.900 --> 00:28:12.300
On to the next one i^2013 something very large.
00:28:12.300 --> 00:28:21.000
We will grab the exponent and we will divide it by 4.
00:28:21.000 --> 00:28:27.400
4 goes into 20 5 times and I will get 20 exactly.
00:28:27.400 --> 00:28:33.000
Let us put in our 0 placeholder over the 1 and then bring down the 13.
00:28:33.000 --> 00:28:40.500
4 goes into 13 3 times and we will get 12.
00:28:40.500 --> 00:28:44.700
This one shows that my remainder is 1.
00:28:44.700 --> 00:28:48.900
What does this entire thing simplify down to?
00:28:48.900 --> 00:28:56.000
If I get a remainder of 1, 2, 3 or 0, now I know what it will simplify down into.
00:28:56.000 --> 00:28:59.700
It simplifies down just to (i).
00:28:59.700 --> 00:29:05.400
There are many things you can do to work with these complex numbers but they are often mere things that you have learned before.
00:29:05.400 --> 00:29:11.000
Keep track of this handy method for simplifying powers of (i) that way you can take those much higher powers
00:29:11.000 --> 00:29:14.100
and bring them down to something nice and compact.
00:29:14.100 --> 00:29:16.000
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