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### Complex Numbers

- The imaginary number i is defined as the square root of -1. This can be used to re-write square roots of any negative number.
- A complex number is a number of the form a + bi. Here a is the real part and b the imaginary part.
- To add or subtract complex numbers think of adding like terms.
- To multiply complex numbers think of multiplying using FOIL. Note that any i
^{2}simplifies to -1. - To divide by a complex number, multiply the top and bottom by the complex conjugate of the denominator.
- Higher powers of i can be simplified into an expression that no longer has a power. One method involves dividing the power by four and checking the value of the remainder.
- Remainder of 1 simplifies to i
- Remainder of 2 simplifies to -1
- Remainder of 3 simplifies to â€“i
- Remainder of 0 simplifies to 1

### Complex Numbers

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Complex Numbers
- Imaginary Numbers
- Complex Numbers
- Real Parts
- Imaginary Parts
- Commutative, Associative, and Distributive Properties
- Adding and Subtracting Complex Numbers
- Multiplying Complex Numbers
- Dividing Complex Numbers
- Complex Conjugate
- Simplifying Powers of i
- Shortcut for Simplifying Powers of i
- Example 1
- Example 2
- Example 3
- Example 4

- Intro 0:00
- Objectives 0:06
- Complex Numbers 1:05
- Imaginary Numbers
- Complex Numbers
- Real Parts
- Imaginary Parts
- Commutative, Associative, and Distributive Properties
- Adding and Subtracting Complex Numbers
- Multiplying Complex Numbers
- Dividing Complex Numbers
- Complex Conjugate
- Simplifying Powers of i
- Shortcut for Simplifying Powers of i
- Example 1 21:14
- Example 2 22:15
- Example 3 23:38
- Example 4 26:33

### Algebra 1 Online Course

### Transcription: Complex Numbers

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at complex number.*0002

*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*0007

*such as multiplying them together which will look like multiplying polynomials and rationalizing denominator will look a lot like the division process.*0013

*Here are some other things that we will cover.*0022

*First we will get into a little bit about that the vocabulary from the complex numbers.*0024

*You have heard about imaginary numbers and you will see how that works with complex numbers.*0028

*We will learn about the real and the imaginary part of complex numbers.*0033

*Many of the properties involving these complex numbers are the same properties that you have seen before for real numbers.*0039

*Then we will get into the nuts and bolts on how you can start combining these things.*0046

*That means adding, subtracting, multiplying and dividing our complex numbers.*0051

*Near the very end we will also see some nice handy ways that we can simplify powers of (i)*0056

*so you can always bring them down to the most simplest form.*0061

*An imaginary number (i) is defined as the âˆš-1.*0067

*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.*0075

*By defining it in this way, the âˆš-1 it can actually express the root of any negative number using this imaginary number (i).*0082

*Here is a quick example to show how it works.*0091

*Suppose Iâ€™m looking at the âˆš-7 using some of our properties we can go ahead and break it up into it -1 Ã— 7.*0093

*And then break up the root over each of those parts.*0102

*You will see the definition shows up right there.*0105

*I'm taking the âˆš -1.*0108

*It is that piece that gets turned into an (i) and now I'm representing this number as (i) Ã— âˆš7.*0111

*Think of doing this for many other numbers where you have a negative underneath the square root.*0119

* If I had -23 underneath the square root that will be (i)âˆš23 and a negative comes out as the (i).*0125

*âˆš -16 would be (i) Ã— 4 or just 4i.*0137

*How those imaginary numbers relate to our complex numbers?*0150

*Complex number is a number of the form A + B(i).*0154

*You can see that the imaginary number is sitting right there, right next to that B.*0158

*Those other values A and B are both real numbers.*0164

*We call A the real part of the complex number and B the imaginary part, since it is right next to the imaginary number.*0168

*You can represent a lot of different numbers using this nice complex form.*0175

*Maybe I will have a number like 3 â€“ 5i that would be in a nice complex form.*0180

*I can read off the real and imaginary part.*0186

*We could throw in some fractions and in here as well, so maybe Â½ Ã— 2/3(i).*0189

*We can also have complex numbers like 0 + 3(i).*0196

*We would probably would not leave it like that, something like this would probably we will write as 3(i).*0202

*Notice how we can still say it is in a complex form where the real part is 0.*0207

*The good part about these complex numbers is they will obey most of the usual laws that you are familiar with.*0217

*Our commutative, associative, and distribution, all of those will work with our imaginary numbers as well.*0223

*The things that we could do with real numbers, we can do with imaginary numbers as well.*0230

*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.*0236

*Let us get into how you can start combining these numbers together.*0247

*To add or subtract complex numbers this is a lot like having like terms.*0251

*Be very careful to watch the signs when combining these numbers together.*0257

*To show you how this works I have 2 â€“ 5(i) that complex number + 1 + 6(i).*0262

*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.*0269

*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.*0280

*That is what I'm talking about when I say adding like terms.*0290

*The real ones 2 + 1 would be 3 and the imaginary I have -5(i) -6(i) that will give me -11(i).*0294

*I will write those as 3 -11(i).*0309

*When you are working with subtraction we have to be very careful with your signs.*0314

*Let me show you, suppose I have 3 + 2(i) and now I'm subtracting a second number 4 + 7(i).*0319

*Parentheses will come in handy because I need to distribute this negative sign to both parts of that second complex number.*0334

*I will look at this as 3 + 2(i) â€“ 4 â€“ 7(i) and now I can see the parts and the like terms that I need to go ahead and put together.*0343

*3 â€“ 4 = -1 and 2 â€“ 7(i) = -5(i)*0357

*This would be written as -1 â€“ 5(i)*0369

*Again, combine those like terms.*0373

*To multiply complex numbers think of multiplying together two binomials.*0379

*It looks a lot like the same process.*0384

*When we are multiplying together two binomials we have great way of remembering that we can use the method of foil to accomplish that.*0387

*There is one additional thing you have to remember, when you have (i) ^{2}, we can go ahead and reduce that to -1,*0394

*that seems a little odd but let us see why that works.*0401

*If I'm looking at the âˆš-1 and that is our (i) and suppose I take that and I square it.*0404

*I want to square that square root I will get -1 on the left.*0413

*That reveals that (i) ^{2} = -1.*0419

*Watch for that to show up in the simplification process.*0424

*Let us go ahead and foil out these two complex numbers.*0428

*2 â€“ 5(i) multiplied by 1 â€“ 6(i)*0432

*We will begin by multiplying the first terms 2 Ã— 1 = 2.*0436

*We will move on to those outside terms 2 Ã— -6(i) = -12(i).*0441

*On the inside terms -5(i) and now let us go ahead and do the last terms -5(i) Ã— -6(i),*0453

*minus Ã— minus would be +, 5 Ã— 6 is 30 and now there is my (i) ^{2}.*0465

*We will go along and we will start crunching things down like we normally would do if they were a couple binomials.*0473

*2 â€“ 17(i) now comes this interesting part this (i) ^{2} here is the same as -1.*0480

*We will go ahead and rewrite it as a -1.*0491

*You can see there is more than we can do to simplify this.*0498

*Iâ€™m not only multiplying the 30 Ã— -1, but then I can go ahead and combine it with the two out front.*0501

*2 â€“ 17(i) - 30 and then let us combine these guys that will give us -28 â€“ 17(i).*0508

*This one is good.*0524

*With many of these problems where combining complex numbers you know that you are done*0527

*when we finally get into that nice complex form.*0532

*You can easily see the real and imaginary parts.*0535

*To go ahead and divide complex numbers, we want to multiply the top and the bottom by the complex conjugate.*0542

*The complex conjugate of a number will look pretty much the same but it will be different in sign.*0550

*If I'm looking at the complex conjugate of A + B(i) that will be â€“B(i).*0557

*Let us just do some quick example.*0563

*Let us see if I have 3 + 2(i) its complex conjugate would be 3 â€“ 2(i).*0564

*If I had -7 â€“ 3(i), the complex conjugate would be -3 + 3(i).*0573

*The only thing that is changing is that negative sign or that sign on the imaginary part.*0583

*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.*0591

* Iâ€™m going to look at the bottom of this particular division problem 4 + 2(i) Ã· 3 + 5(i).*0599

*I'm going to find the complex conjugate of 3 + 5(i), that would be 3 â€“ 5(i).*0606

*Once we find that complex conjugate we will multiply on the top and on the bottom of our complex fraction.*0616

*Multiply on top, multiply on the bottom.*0628

*This one involves quite a bit of work.*0631

*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.*0634

*This will make things look a lot more complicated at first but if you just go through the problem carefully you should do fine.*0644

*Let us start with the top.*0653

*Taking our first terms I have 4 Ã— 3 and that would give me 12.*0656

*The outside terms would be 4 Ã— -5 (i) = -20(i).*0662

*On the inside 2(i) + 3(i) then would be 6(i) and our last terms 2 Ã— -5 = -10(i) ^{2}.*0668

*Do not forget to multiply those (i) together as well.*0680

*On the button 3 Ã— 3 = 9, outside will be -15(i), inside would be 15(i).*0684

*And then we have 5(i) Ã— -5(i) = -25(i) ^{2}.*0695

*Some things you want to notice after you go through that forming process.*0703

*We are using that complex conjugate on the bottom, the outside and inside terms will end up canceling.*0707

*If they do not cancel, make sure you have chosen the correct complex conjugate.*0715

*Notice that we have a couple of (i) ^{2} showing up on the top and bottom.*0720

*That comes from the (i) being multiplied together.*0725

*Each of these will be need to change into -1.*0728

*Let us go through and do those two things and clean this problem a little bit.*0733

*I have 12 â€“ 20 + 6(i) = -14(i), -10 Ã— -1 Ã· 9 my outside and inside terms on the bottom cancel -25 Ã— -1.*0737

*We can see the beauty of using that complex conjugate.*0759

*On the bottom we no longer have any more imaginary numbers.*0762

*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.*0766

*Here we got rid of the imaginary numbers in the bottom.*0772

*Let us continue to simplify and see where we can take this problem.*0776

*12 â€“ 14(i) + 10 Ã· (9 + 25)*0781

*12 + 10 = 22 â€“ 14(i) Ã· 34*0790

*It is tempting to try and stop right there but do not do it, we want to continue doing this entire problem*0801

*and try to get it in that nice complex form where I can see the real and imaginary part.*0806

*To get in that form Iâ€™m going to use the part where I breakup the 34 and 22 and under the 14.*0811

*It looks a lot like when we are taking polynomials when we are dividing by a monomial.*0818

*This will be 22 Ã· 34 -14(i) Ã· 34.*0824

*This particular one, both of those fractions can be reduced.*0833

*I will divide the top and the bottom by 2, 11/17 â€“ 7(i) Ã· 17.*0837

*This one is actually completely done.*0846

*I can easily see my real part over here and my imaginary part -7/17.*0849

*That is I know that this one is done.*0855

*Division is quite a tricky process remember to multiply by the complex conjugate of the bottom*0859

*and then work very carefully to simplify the problem from there.*0864

*It should be in its complex form when you are all done so, you can see the real and imaginary parts.*0868

*After working with doing some combining things with these various complex imaginary numbers,*0876

*you may have noticed that you rarely see any higher powers of (i).*0881

*In fact the largest part of (i) that we came across was i ^{2} and we turned it immediately into a -1.*0885

*The interesting part is you can usually take much higher powers of (i) and end up simplifying them down.*0892

*To see out why this works for some much higher powers, let us just take a bunch and start picking them apart.*0898

*We see many instances while we are working with (i) and (i) was just equal âˆš1 which of course we call (i).*0905

*When we have i ^{2} and then we thought about squaring the âˆš-1, so we got -1 as its most simplified result.*0912

*We will see what will happen with i ^{3}.*0922

*One way that you could interpret this would be i ^{2} Ã— (i).*0927

*The reason for doing that is because you can then take the i ^{2} sitting right here and simplify that.*0934

*That will give us -1 Ã— (i) the final result of that would be â€“i as it is simplest form.*0942

*Let us choose that same idea and see if we can go ahead and simplify i ^{4}.*0955

*That will be i ^{2} Ã— i^{2}.*0959

*Both of the i ^{2} = -1*0964

*I have -1 Ã— -1 = 1*0970

*Take note how all of these are simplifying to something else that does not have any more powers on them.*0975

*Let us do a few more that hopefully we can develop a much easier or a shortcut way of doing this process.*0983

*Let us go on to i ^{5}.*0990

*This can be thought of as i ^{4} Ã— (i).*0994

*We have already done i ^{4}, it is way back here, that is equal to just 1.*0999

*1 Ã— (i) I know this simplifies to (i).*1003

*On the next one i ^{6}, i^{4} Ã— i^{2}.*1010

*We have already done i ^{4} that is just 1.*1020

*I will put in i ^{2}, I will just write across from it -1 so 1 Ã— -1 = -1, very interesting.*1024

*I think we are starting to see a pattern of some of these (i).*1035

*Notice how they are looking exactly the same as the ones on the other side here.*1038

*Let us see if that continues.*1044

*I ^{7} will be i^{4} Ã— i^{3}, 1 Ã— i^{3} = -i.*1046

*Maybe the last one, i ^{4} Ã— i^{4}, both of those are equal to one, this is equal to 1 as well.*1058

*Notice how we did starting to see same numbers.*1069

*There are some good patterns we can see, I mean we got past i ^{4} and all of them contained an i^{4}*1072

*so we are getting this similar pieces right here.*1079

*If I was even to continue on to i ^{9} then I will have even more groups of i^{4} and I could see that it would simplified down.*1083

*All of those much higher powers will end up simplifying down to something which does not have any more powers whatsoever.*1097

*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).*1104

*Let us see how to do that.*1111

*Looking at all of these examples you can see that there is a pattern on how they simplify.*1115

*In fact they repeats in blocks of 4 and you start with (i) then it goes -1, then â€“I, then 1, it just repeats over and over again.*1120

*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.*1129

*From the remainder you will know exactly what it will simplify to.*1138

*I have made a handy table to figure out what that looks like*1142

*If Iâ€™m looking at (i) to some large power, I will take the power and divide it by 4*1146

*and the remainder happens to be 3, according to my chart here it will simplify to â€“i.*1152

*Let us do a quick example to see how this works and why it works.*1160

*Let us take i ^{13}.*1166

*If I wanted to I could imagine a whole bunch of i ^{4} in here.*1173

*i ^{4}, i^{4}, i^{4} and then i^{1}.*1178

*If we add up all those exponents they add up to 13.*1186

*You can see that most of them are gone since i ^{4} = 1.*1191

*1 Ã— 1 Ã— 1 Ã— (i) =(i)*1196

*Let us use our shortcut formula to figure out the same thing.*1207

*In the shortcut we take the exponent and divide that by 4.*1210

*13 Ã· 4, 4 goes in the 13 exactly 3 times.*1216

*Notice why that is important, that is exactly how many bunches of i ^{4} I have.*1227

*When you are going through that division by 4 process you are counting up all of the i ^{4} that will simply break down and simplify into 1.*1233

*It goes in there 3 times, 3 Ã— 4 = 12 and I will subtract that away getting my remainder of 1.*1243

*That remainder tells me how many extra (i) I still have left over.*1253

*From there I can see I have only one (i) and I know it simplifies down to (i).*1259

*I get the same exact answer as the 4, i ^{13} = (i).*1267

*Now that we know a lot more about these complex numbers let us go through some various examples*1275

*of putting them together and see how we can simplify powers of (i).*1279

*This first one we want to do a subtraction problem 4 + 5(i) â€“ 6 â€“ 3(i).*1284

*With the addition and subtraction we are looking to add like terms.*1290

*Subtraction is a tricky one, you have to remember to distribute our sign onto both parts of that second complex number.*1295

*Iâ€™m looking at 4 + 5(i) -6 + 3(i)*1302

*I can see my like terms, let us put together 4 and -6, and 5 and 3.*1311

*4 - 6 = -2, 5(i) + 3(i) = 8(i)*1320

*We can see that it is in that final complex form and we know that this one is done.*1329

*In this next example, we want to multiply two complex numbers.*1337

*Think foil.*1342

*Be on the watch out for additional things that we will need to simplify such as the i ^{2}.*1346

*Okay, starting off, multiplying our first terms together we will get 1.*1351

*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) Ã— 3(i) + 6i ^{2}.*1357

*We go through and start combining everything we can.*1377

*1 + I have my like terms here, so 3 + 2 = 5(i) + 6 and take the i ^{2} and turn it into -1.*1380

*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.*1397

*This will be -5 + 5(i) and now I can easily see my real and imaginary part, so I know that one is done.*1408

*Let us divide these two complex numbers.*1420

*I have 2 â€“ 5(i) Ã· 1 â€“ 6(i)*1422

*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.*1426

*Let us first find that complex conjugate.*1434

*That will be 1 + 6(i).*1437

*We are going to use that on the top and on the bottom.*1440

*We have to remember that since we are multiplying we will have to foil out the top and the bottom.*1445

*Foil top and foil bottom.*1452

*Let us see what the result of that one looks like.*1458

*Iâ€™m looking at the top, first terms 2 Ã— 1 =2, outside terms would give me 12(i), inside terms -5(i), and last terms -30(i) ^{2}.*1461

*Lots of terms in there, be very careful and keep track of them all.*1479

*On the bottom, first terms 1, outside 6i, inside -6i, and our last terms -36i ^{2}.*1483

*It is looking much better, looking pretty good.*1496

*If we do this correctly, we should not have any more powers of (i) in the bottom.*1498

*Let us see what else we can simplify.*1502

*We will go ahead and combine these 2i on the top and we will combine these 2i of the bottom*1505

*and then we will put these i ^{2} to make them both -1.*1510

*2 and then we will have 7(i) - 30 Ã— -1.*1518

*On the bottom 1 â€“ 36 -1.*1527

*It looks like we have indeed accomplished our goal there is no longer imaginary numbers on the bottom.*1533

*We continue putting things together 2 + 7(i) + 30 since the negative Ã— negative is a positive and 1 + 36.*1538

*Just a few more things to put together 32 + 7(i) Ã· 37.*1553

*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.*1562

*Iâ€™m going to give 37 ^{32} and 37^{7}.*1570

*7 /37(i) this one looks much better.*1579

*Sometimes you can go ahead and reduce those fractions but this one the way it is.*1587

*In this last problem we will go ahead and see if we can simplify some very large powers of (i).*1595

*Since they are very large powers we will go ahead and use our shortcut to take care of that process.*1602

*In the first one I have i ^{107}.*1607

*Let us start off by taking that exponent, 107 and we will divide it by 4.*1611

*This will figure out all the bunches of i ^{4} that we have hiding in there.*1618

*4 of those in the 10, twice and we get 8.*1623

*Subtract them away I will end up with 27.*1629

*That goes in there 6 times and 6 Ã— 4 = 24 and then we can subtract that away.*1635

*I'm down to 3 and that is smaller than 4 so I know that 3 is my remainder.*1645

*Think back, if my remainder is 3 then what does this entire thing simplify to?*1653

*It simplifies to â€“i.*1660

*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).*1664

*(i) = (i), i ^{2} = -1, and we have â€“I and 1.*1674

*You quickly have built your table.*1682

*On to the next one i ^{2013} something very large.*1686

*We will grab the exponent and we will divide it by 4.*1692

*4 goes into 20 5 times and I will get 20 exactly.*1701

*Let us put in our 0 placeholder over the 1 and then bring down the 13.*1707

*4 goes into 13 3 times and we will get 12.*1713

*This one shows that my remainder is 1.*1720

*What does this entire thing simplify down to?*1725

*If I get a remainder of 1, 2, 3 or 0, now I know what it will simplify down into.*1729

*It simplifies down just to (i).*1736

*There are many things you can do to work with these complex numbers but they are often mere things that you have learned before.*1740

*Keep track of this handy method for simplifying powers of (i) that way you can take those much higher powers*1745

*and bring them down to something nice and compact.*1751

*Thank you for watching www.educator.com.*1754

0 answers

Post by David Saver on August 13, 2014

Great Course!

3 answers

Last reply by: Professor Eric Smith

Sun Jun 8, 2014 9:26 PM

Post by Mohamed Elnaklawi on April 6, 2014

what does FOIL stand for?

1 answer

Last reply by: Professor Eric Smith

Mon Dec 2, 2013 8:50 PM

Post by Emily Engle on December 1, 2013

Thanks for your lectures; they are very helpful. Are you going to do algebra 2?