For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

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### Complex Numbers

- So far, we've found that some polynomials are
*irreducible*in the real numbers: they cannot be broken down into smaller factors and they have no roots. For example, x^{2}+1 has no roots and cannot be broken down any further. - We introduce a new number to get around this: the
*imaginary number*,

Notice that i

√−1

= i. ^{2}= −1, that is, we can square it and get a negative. That's something we could never do in the real numbers! - With this idea in mind, we can now take the square root of any negative number. The negative just pops out as i, while we do the rest of the square root normally. For example, √{−49} = 7i.
- With the idea of the imaginary number i in place, we can create a new set of numbers, which we'll call the
*complex numbers*(denoted by ℂ). We still want to be able to talk about real numbers (ℝ), so we'll need them to appear along with i. Thus we'll make each complex number with a*real part*and an*imaginary part*:

where a and b are both real numbers. [a is the real part, bi is the imaginary part.]a + b i, - Working with complex numbers is similar to working with real numbers.
- Two complex numbers are equal when the parts in one number are equal to the parts in the other.
- We can do addition and subtraction by having the real parts add/subtract together and the imaginary parts add/subtract together. Other than the two parts staying separate, it works like normal. Notice how this is similar to adding/subtracting like variables in algebra.
- Multiplication also works very similarly to what we're used to. Just approach it like a FOIL expansion. [Since i
^{2}= −1, it will transform during simplification.] - Division is a little trickier. First, we need the idea of a
*complex conjugate*. The conjugate of (a+bi) is (a−bi) and vice-versa. Notice that whenever you multiply a complex number by its conjugate, it always results in a real number. This means if we have a fraction (division) with a complex number in the denominator, we can multiply the numerator and denominator by the complex conjugate for the denominator so that we can now have a real in the denominator. Then we just divide like usual.

- Now that we understand how complex numbers work, we can revisit the quadratic formula and use it to find the roots of "irreducible" quadratics.

Previously, we couldn't use the formula if bx = −b ±

√b^{2}−4ac

2a

^{2}− 4ac < 0, but now we know it just produces an imaginary number! Furthermore, because of the ±√{b^{2}−4ac} part in the quadratic formula, we see that complex roots must come in conjugate pairs. That is, if p(x) is a polynomial:p(a+bi) = 0 ⇔ p(a−bi) = 0.

### 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
- Introduction
- A Wacky Idea
- Square Roots and Imaginary Numbers
- Complex Numbers
- Addition and Subtraction
- Multiplication
- Division
- Complex Conjugates
- Division through Complex Conjugates
- Factoring So-Called 'Irreducible' Quadratics
- But Are the Complex Numbers 'Real'?
- Still, We Won't See Much of C
- Example 1
- Example 2
- Example 3
- Example 4

- Intro 0:00
- Introduction 0:04
- A Wacky Idea 1:02
- The Definition of the Imaginary Number
- How it Helps Solve Equations
- Square Roots and Imaginary Numbers 3:15
- Complex Numbers 5:00
- Real Part and Imaginary Part
- When Two Complex Numbers are Equal
- Addition and Subtraction 6:40
- Deal with Real and Imaginary Parts Separately
- Two Quick Examples
- Multiplication 9:07
- FOIL Expansion
- Note What Happens to the Square of the Imaginary Number
- Two Quick Examples
- Division 11:27
- Complex Conjugates 13:37
- Getting Rid of i
- How to Denote the Conjugate
- Division through Complex Conjugates 16:11
- Multiply by the Conjugate of the Denominator
- Example
- Factoring So-Called 'Irreducible' Quadratics 19:24
- Revisiting the Quadratic Formula
- Conjugate Pairs
- But Are the Complex Numbers 'Real'? 21:27
- What Makes a Number Legitimate
- Where Complex Numbers are Used
- Still, We Won't See Much of C 29:05
- Example 1 30:30
- Example 2 33:15
- Example 3 38:12
- Example 4 42:07

### Math Analysis Online

### Transcription: Complex Numbers

*Hi--welcome back to Educator.com.*0000

*Today, we are going to talk about complex numbers.*0002

*At this point, we know a lot about factoring polynomials and finding their roots.*0005

*Still, there are some polynomials we can't factor.*0009

*There is no way to reduce them into smaller factors, so they are irreducible; and they simply have no roots.*0011

*There is just no way to solve something like x ^{2} + 1 = 0.*0016

*There are no roots to x ^{2} + 1, because for that to be true, we would need x^{2} = -1.*0020

* But when you square any number, it becomes a positive; if we square +2, then that becomes +4.*0025

*But if we square -2, then that is going to become +4, as well; the negatives hit each other, and they cancel out.*0033

*This pattern is going to happen for any negative number and any positive number, and 0's are just going to stay 0's.*0040

*So, there is no way we can square a number and have it become negative.*0045

*But what if that wasn't the whole story--what if there was some special number we hadn't seen, that, when squared, does not become positive?*0050

*That is an interesting idea; if that is the case, we had better explore--let's go!*0058

*We are looking for a way to solve x ^{2} = -1; in other words, we are looking for something that is the square root of -1.*0063

*We are looking for something that, when you square it, gives out -1.*0070

*So, here is a crazy idea: why don't we just make up a new number?*0074

*We will try something really crazy, and we will just create a number out of whole cloth.*0078

*We will imagine a special number that becomes -1 when squared; so we are making a new number.*0082

*And since we are using our imagination to think of this new thing, we will call it an imaginary number, and we will denote it with the symbol i.*0088

*We know that i = √-1, whatever that means...which means that, when you square i, you get -1.*0097

*The square root of 4 is 2, because when you square 2, you get 4.*0104

*So, when you square i (since it is the square root of -1), you get -1.*0108

*These two ideas are how we are going to define this new thing.*0113

*When you are writing it, I would also recommend writing it with a little curve on the bottom, just so you don't get it confused.*0117

*Sometimes, if you are writing quickly, you might end up not even putting a dot,*0122

*at which point it would be hard to see whether you meant to put down 1 or you meant to put down i.*0125

*So, I recommend putting a tiny little curve at the bottom, and then a dot like that, when you are writing it by hand.*0130

*That way, you will have some way of being able to clearly see that you are talking about the imaginary number, and not a normal number.*0134

*With this new idea of i, we can solve the original equation: since i is equal to √-1, we can just take the square root of both sides.*0141

*And remember: when you take the square root of both sides, you have to introduce a plus/minus.*0148

*The square root of both sides of an equation...plus/minus shows up, no matter what.*0152

*So, we get x = ±i; let's check it really quickly.*0155

*If we have positive i, squared, then that is going to be equal to i ^{2}, which is equal to -1, as we just had here.*0159

*And our other possibility, if we have (-i) ^{2}...well, negative times negative becomes positive.*0168

*So, we have positive i squared; but i ^{2} is, once again, -1; so it checks out--both of these are, indeed, solutions.*0174

*So, we have found how to solve x ^{2} = -1.*0183

*We have this new idea of taking a square root of -1, which means, when we square that thing,*0186

*this thing that we have just created, the imaginary number, we will have -1.*0191

*We can now use this idea to take the square root of any negative number.*0197

*We are not just limited to taking the square root of -1; we can do it on anything.*0199

*We just separate out that √-1; that will become an i; and then we take the root as normal.*0203

*For example, √-25 would become...we pull out -1, so that is the same thing as 25 times -1.*0207

*So, we separate this out; we break it into two square roots (a rule we are allowed to do): times √-1.*0218

*So, √-1 becomes i; √25 is 5; so we have gotten 5i out of that.*0223

*We look at the square root of -98; well, that is the square root of 98, times -1 inside.*0230

*So, we can separate that out, and we will get √98 times √-1.*0236

*√-1 will become i; but what is the square root of 98?*0242

*I am not quite sure, so we need to break it up a bit more; look at how we can break that into its multiplicative factors.*0245

*98 is the same thing as 49 times 2; the square root of 49 is 7; 7 times 7 is 49; so we have 7√2i.*0250

*Finally, the square root of -60; we can see that as 60 times -1, so we separate out the -1 from √60, so this becomes i.*0260

*What is √60? How can we break that into its factors?*0270

*Well, we have a 6 times a 10; that is still not quite enough, though, so we break that up some more.*0273

*We have 2 times 3 for 6, and 2 times 5 for 10, still times i; we see we have a 2 here and a 2 here,*0279

*so we can pull them out, because they come as a pair.*0286

*2 times √(3 times 5); we can't do that, so we might as well just turn it into one number, 2√15 times i.*0288

*So, we can now take the square root of any negative number by having this idea of i, the imaginary number.*0295

*With this idea of the imaginary number in place, we can create a new set of numbers, which we will call the complex numbers.*0302

*And we denote it with ℂ; if you are writing that by hand, you make a normal C, and then you make a little vertical line like that.*0307

*We still want to be able to talk about the real numbers, which, remember, we denote with this weird ℝ symbol.*0313

*So, we will need them to appear along with i; so we need to have real numbers show up along with this imaginary number.*0318

*And so, we just saw that we can have imaginary numbers that had this real component multiplied against them.*0323

*We took the square root of -25, and we got 5i; so we are going to have to have some number times i;*0328

*and we will also want to have a real part, a; so we will have the real part that will be a.*0335

*a is the real part; and then we will have the imaginary part that will be bi; bi is the imaginary part.*0343

*And a and b are both real numbers; they are just coming out of that real set, like usual.*0353

*This gives us an entirely new form of number: as opposed to just being stuck with real numbers,*0358

*we have a way of having a real number and a complex number, both interacting with each other.*0362

*They will come together as a package; and this is the idea of the complex numbers.*0366

*Two complex numbers are equal when the parts in one number are equal to the parts in the other.*0371

*If we have a + bi, and we are told that that is equal to c + di, then that means the real parts have to be equal.*0375

*So, we know that a and c are equal.*0383

*Also, we know that the imaginary parts have to be equal; both parts have to be equal for this equality to hold.*0387

*bi and di are the same thing, which means that b and d must be the same thing, since clearly i is going to be the same thing on both sides.*0393

*Great; all right, so how do we do our basic arithmetic with these things?*0399

*Addition and subtraction first: we add and subtract complex numbers pretty much like we are used to.*0404

*a + bi plus c + di just means we are going to combine our real parts (they will become a + c);*0408

*and our imaginary parts (bi and di) will come together, and we will get (b + d) all times i; bi + di becomes (b + d)i.*0415

*The same thing over here; if we have a + bi minus c + di, then we have (a - c), and the bi's do the same thing with the di's.*0426

*So, we are going to have b - d, because remember: there is that minus symbol there; so it is b - d there.*0439

*a + bi minus c + di...we can also think of that as just distributing this negative sign, like we did before.*0445

*So, we are now adding a -c and a -di; that is one way of looking at subtraction.*0451

*So, real parts add and subtract together, and imaginary parts add and subtract together.*0456

*Other than the fact that they stay separate, it pretty much works like normal.*0460

*So, as long as they stay separate--they stay on their two sides--imaginaries can't interact directly with the reals;*0464

*reals can't interact directly with the imaginaries when we are keeping it in addition and subtraction--it is pretty much just normal.*0469

*Let's look at two examples: 5 + 2i plus 8 - 4i--our 5 and 8 will be able to interact together, because they are both real numbers.*0475

*And let's color-code that, just so we can see exactly what is going on.*0483

*So, with our colors before, with red representing reals again, we have 5 + 8, and then it will be +, and then our imaginaries interact as well; 2i - 4i.*0486

*5 + 8...we get 13; and 2i - 4i (because this was a -4i here)...we get -2i, which we could also write as 13 - 2i.*0497

*13 + -2i and 13 - 2i mean the same thing; great.*0511

*If we did it with subtraction, (5 + 2i) - (8 - 4i), then we can distribute this negative sign; and we get -8,*0516

*and then we will cancel out the minus there, and we get + 4i; so 5 - 8 + 2i + 4i.*0524

*5 - 8 becomes -3; plus 6i; great.*0540

*All right, multiplication is pretty much going to work very similarly, as well.*0547

*Now, notice: we have two things in this sort of factor-looking form; so we do it as a FOIL expansion.*0551

*We are going to do that same idea of distribution.*0558

*We will multiply again in the same way that we distributed before.*0559

*a will multiply on c, and a will multiply on di; so we get ac and adi.*0563

*And then, bi will multiply on c, and bi will multiply on di; so we will get...bi on c gets us bci, and bi on di will get us bdi ^{2}.*0570

*Now, remember: i ^{2} = -1; so when we have this i^{2} here, it cancels out, and it is like we have subtraction.*0581

*So, we put together our real components now: bd and ac gets us ac - bd, because it was -bd.*0590

*And we have our imaginary components, adi and bci, so we have (ad + bc)i.*0599

*Remember, this i ^{2} equals -1, so it will just transform during the process of simplification.*0606

*Now, you could memorize this formula right here, but I wouldn't recommend memorizing this formula right here,*0611

*because you already know the FOIL expansion, and as long as you can remember i ^{2} = -1, that will keep it easier.*0616

*That is the better way to do it.*0621

*All right, let's see some examples: (1 + 2i)(5 - i): 1 times 5 becomes 5; 1 times -i becomes -i;*0623

*2i times 5 becomes 10i; 2i times -i becomes -2...and we have two i's, so 2i ^{2}.*0632

*Once again, i ^{2} is equal to -1; that cancels out and becomes a plus.*0639

*So, 5 - i + 10i + 2: 5 and 2 combine to give us 7; and -i + 10i combine to give us + 9i.*0645

*Great; the next one is (6 + 10i)(5 + 3i): 6 times 5 is 30; 6 times 3i is 18i; 10i times 5 is 50i; and 10i times 3i is 30i ^{2}.*0658

*Once again, remember: i ^{2} is -1, so we have -30, at which point our -30 and positive 30 cancel each other out.*0674

*And we are left with just 18i + 50i, so we get a total of 68i.*0681

*Great; the final one is division; now, division is a little more tricky.*0686

*Consider if we had (10 - 15i)/(1 + 2i): now, at first, we might think,*0690

*"Oh, we have 10 and 1; we have -15 and 2i; so we will get 10/1 and -15/2,*0695

*because the i's will cancel out"; but that would be wrong.*0702

*We have to divide by the entire denominator, not just bits and pieces.*0704

*For example, to see why this has to be the case, imagine if we had 5 + 5, over 3 + 2; that is really 10/5, which equals 2.*0708

*But we could get confused and think that that was going to be 5/3 + 5/2; but division does not distribute like that.*0719

*We are not allowed to do that; so we can't do the same thing here with our 1 + 2i.*0729

*We can't break it up and distribute the pieces, because it is nonsense in real numbers; so it is definitely going to be nonsense in the complex.*0733

*What if we break it up, and we put 1 + 2i onto the 10, and then we put 1 + 2i on the -15i separately?*0740

*We get 10/(1 + 2i), and we get + -15/(1 + 2i).*0747

*Well, that is true; we broke it up; we can do that with normal things.*0754

*We could have it, if we wanted to, going back to 5 + 5 over 3 + 2--we could have that as 5/(3 + 2) + 5/(3 + 2).*0758

*But that doesn't help us; we still have to divide, ultimately, by this 1 + 2i.*0766

*We don't know how to divide by a complex number yet; that is our problem.*0774

*Simply put, we have no idea how to divide by complex numbers; that is our problem.*0779

*Addition and subtraction made natural sense; real numbers stuck together; imaginary numbers stuck together.*0786

*FOIL was able to allow us to do multiplication--we just did normal distribution, and we remembered the rule that i ^{2} becomes -1.*0791

*But division...we don't have a good understanding of what it means to divide by a complex number.*0797

*That is tough; now, what we could do, if there was some clever way to get rid of having a complex number in the denominator--*0803

*if we could somehow make it into an alternate form where we disappeared the complex number in the denominator-- we would be good.*0810

*Hmm...to figure out this clever method that we want, first notice something you might have seen while we were working on quadratics.*0818

*If you have (x - 2)(x + 2), you get x ^{2}, and then + 2x, but also - 2x;*0824

*since we have the -2 and the +2 here, they end up canceling each other out, and so we are left with just x ^{2} - 4.*0830

*And there is no middle term with just x; there is no x that shows up.*0837

*We are able to get rid of it, and have only the doubled and then no x whatsoever.*0843

*We can expand this idea to complex numbers: we do a similar pattern, (1 - 2i)(1 + 2i); let's work that out.*0848

*We get 1 - 2i + 2i - 4i ^{2}; so +2i and -2i cancel each other out, because we have the negative here and the plus here.*0855

*And then, we have 4i ^{2}, so that will cancel and become a plus; and we get 1 + 4 = 5.*0864

*So, we have been able to figure out a way to multiply this thing and get just a real number.*0870

*So, if we use this pattern, (a + bi)(a - bi), you automatically get a real number.*0875

*When you multiply it out, it results in a number that has no imaginary part.*0881

*You can get that i to disappear entirely, and get something that is completely real.*0884

*This idea we call the complex conjugate; it comes up often enough, and it becomes important enough,*0890

*that we give it a special name (complex conjugate); and we also give it a special symbol.*0895

*We denote it with a bar over the number; so if we have a + bi as our complex number,*0899

*we can talk about its conjugate with a bar over it; and that is a - bi.*0904

*The conjugate of a + bi is a - bi; and what is the conjugate of a - bi?*0909

*Well, we just flip it again, back to a plus, vice versa; a + bi is the conjugate of a - bi; the conjugate of a - bi is a + bi.*0913

*They just end up flipping between each other, as long as we are doing conjugates.*0920

*Notice that, whenever you multiply a complex number by its conjugate, it always results in a real number.*0924

*So, by multiplying with the conjugate, we can get rid of imaginary things; we can get rid of it.*0930

*Multiply by a conjugate; you always get a real number out of it.*0937

*So, let's look at that: a + bi times a - bi: we get the a ^{2}, and we will get - abi + abi;*0941

*plus here, minus here; those cancel out; i ^{2} is a -1, so that causes that to become a plus.*0948

*So, we will end up with a ^{2} + b^{2}, and no i; a and b were just real numbers, so we have something that is entirely real.*0955

*We started with imaginary things, but by multiplying these together,*0963

*choosing carefully what we had, we were able to knock out imaginaries entirely and get something that is just real.*0966

*With this idea of complex conjugates in mind, we can now deal with division.*0972

*We simply turn the denominator into a real number by multiplying top and bottom of the denominator's conjugate.*0976

*We want to get the bottom to turn into just a real number, because we know how to divide by reals; you just put it on a fraction.*0981

*So, with c + di, we need to multiply c - di.*0987

*Of course, we can't just multiply the bottom because we feel like it;*0991

*so we also have to multiply the top by c - di, as well, because something over itself is always 1*0994

*(as long as you didn't start with 0/0; then the world explodes).*1000

*But as long as it is something over something, and that something isn't 0, you get 1.*1003

*So, c - di over c - di...we can do that; we just trust, intrinsically, that dividing something by itself is 1.*1006

*That is the nature of division; that is the point of it.*1012

*So, ac + bd + bc - ad times i over c ^{2} + d^{2}; that is what it will end up simplifying to.*1015

*And we could work this out, and we would see that this formula ends up working out.*1023

*I don't want you to memorize this formula; I don't even really see a good point to working through it.*1027

*The important thing to know is: just remember to multiply top and bottom.*1030

*Remember to multiply top and bottom by the denominator's conjugate.*1034

*This idea of being able to multiply by a conjugate--that is the really cool thing.*1039

*You could memorize a formula, but it is not going to help you to memorize a formula, because it is hard to recall a formula like this.*1043

*It is much easier to remember that I have division of a complex number.*1050

*I multiply by the conjugate, because I want to get rid of real numbers.*1054

*I have to multiply top and bottom, though, because of course, if you did otherwise, you would just be playing fantasy.*1057

*You have to multiply by the same thing on the top and the bottom to keep it what you started with.*1061

*All right, let's see an example: (10 - 59) over (1 + 2i).*1064

*We want to multiply by the conjugate of (1 + 2i) (if we wanted to, we could express that with a bar all over the top of it), which would be (1 - 2i).*1070

*We multiply by that, and we know we will have gotten to just a real number.*1079

*1 - 2i multiplies top and bottom; and it does have to come in parentheses, because it is a whole thing multiplying some other whole thing.*1083

*You don't just get multiplied bits and pieces.*1090

*We work that out: 10 times 1 gets us 10; 10 times -2i gets us -20i; -15i times 1 gets us -15i; -15i times -2i gets us +30i ^{2}.*1093

*What is on the bottom? We have (1 + 2i)(1 - 2i); 1 times 1 gets us 1; + 2i, - 2i; those will cancel out; -2i + 2i, and then -4i ^{2}.*1107

*Remember, i ^{2} becomes -1; so we cancel out like that.*1126

*And then, we also see that -2i + 2i cancel each other out.*1132

*What does this become next? We combine things: 10 - 30, our real parts on the top, become -20.*1135

*-20i - 15i becomes -35i; what is on the bottom? 1 + 4 is 5, so we can divide -20/5, minus 35i/5; so that gets us -4 - 7i.*1141

*Great; all right, now that we understand the basics of how to work with complex numbers,*1162

*we are now at a point where we can actually see how to factor irreducible quadratics.*1168

*It is now possible for us to factor previously irreducible quadratics and find their roots.*1172

*So, x ^{2} + 1, we see, is now factored into (x + i) and (x - i).*1176

*Let's check this: we get that this would be equal to x ^{2} - ix + ix - i^{2}.*1180

*Oops, ix; I didn't write that whole thing.*1189

*Those cancel each other out; i ^{2} becomes + 1, so we get x^{2} + 1.*1193

*Sure enough, it checks out; and we have found a way to be able to factor this thing that, before, we could not factor.*1201

*It used to be irreducible, but now we see that, through the complex numbers, it is not irreducible at all.*1206

*It is totally factorable; we can revisit the quadratic formula and use it to find the roots of these supposedly irreducible quadratics.*1210

*What used to be irreducible for us is no longer, so we can use the quadratic formula.*1216

*Previously, we couldn't use it when b ^{2} - 4a was less than 0, because there was no square root of a negative number.*1221

*But now we know that that just means an imaginary number; so if our discriminant, b ^{2} - 4a, shows it is less than 0,*1226

*then that means, not that we have no answers, but that we just have imaginary answers.*1232

*Cool; furthermore, because of the ± √(b ^{2} - 4ac) part in the quadratic formula,*1236

*we see that complex conjugates must come in conjugate pairs.*1242

*If b ^{2} - 4ac was less than 0, so this gives out stuff times i, then we have this ± thing;*1246

*so it is going to be plus stuff(i), minus stuff(i); so we have one version that is a +i and one version that is a -i.*1254

*That is what happens when we are doing a conjugate pair.*1260

*We have a + bi; its conjugate is a - bi, so if we have stuff + stuff(i) and minus stuff(i), that is what we have right there.*1264

*All right, so if we have a polynomial where we know that a + bi is a root*1274

*(that is to say, when you plug it in you get 0), then we know that a - bi has to also be a root;*1279

*these things come in conjugate pairs all the time.*1285

*So, we talked a lot about the complex numbers; but we probably have this nagging question in the back of our head.*1288

*Are they real? They are clearly not real numbers, because we are saying that they are not the real numbers,*1294

*which are numbers like 5, 0, π, √2...we have been working with them all up until now.*1299

*But are they real--are they legitimate--are they something that we really can use,*1306

*and not be thinking that we shouldn't be using these?*1310

*I mean, they have the word "imaginary" in their definition; do we really want to be trying to do science or math with something that is inherently imaginary?*1313

*Let's think about this: what does it mean for a number to be legitimate?*1323

*What is this idea of a number being a legitimate number that is valid for science, valid for math?*1327

*Now, we probably all agree that 1, 2, 3...those are totally valid.*1333

*You could pick up one rock; you could pick up two rocks; you could pick up three rocks.*1337

*We could actually have these things in our hands and say, "Look, I have that many objects."*1341

*And we might not be able to pick up 5 billion rocks, but we can get this idea that we could count that many rocks in front of us.*1346

*So, that seems pretty valid; these nice whole numbers are perfectly reasonable.*1352

*But what about 1/2 being a real number? 1/2 seems pretty valid, because we could take a pizza, and we could cut the pizza in half.*1357

*We take a pizza; we cut it down the middle; and now, all of a sudden, we have two chunks of pizza.*1366

*1 here; 1 here; we are left with two objects that come together to form a whole.*1372

*But at the same time, we could say, "Well, this is one object, and this is one object; so it is 1 and 1."*1378

*But we could also say it is 1/2 of what we originally started with, and it is 1/2 of what we originally started with.*1384

*So, it is 1/2 and 1/2; so it is a little bit more questionable that this is valid,*1390

*because can you actually pick up a half-object? No, it is an object in and of itself.*1397

*But it is connected to other things, so it is not perfectly valid--not as valid as the rocks.*1401

*We can grab rocks; we can hold rocks; but we can definitely believe in half-numbers.*1405

*We can believe in rational numbers; we can believe in fractions; it seems reasonable.*1410

*Well, OK, what about something even more slippery--what about the negative numbers?*1414

*Negatives are pretty bad; or we could go even worse, and we could talk about irrational numbers.*1419

*How can you possibly hold -1 rock? What does that mean?*1425

*Can you hold √2 rock? Can you hold π rock?*1428

*You can't hold these things in your hand; so are they valid?*1431

*We can't cut a π slice of pizza; we can't cut a √2 slice of pizza; what does this mean now?*1433

*Are they really valid? We certainly used them a lot before--we are used to using them.*1440

*So, they seem reasonable in that way; but are they things that are real in the real world?*1445

*Don't worry; they are not illegitimate; it doesn't mean that they are illegitimate because you can't hold them in your hand.*1451

*We can use them to represent things in the real world.*1456

*We can talk about an object falling with negative numbers; we can say it is going a negative height.*1458

*As opposed to a positive height, where it goes up, it goes a negative height, where it goes down.*1464

*Or maybe you have $100 in your bank account, but then you pull out 150; that leaves you with negative $50 in your bank account.*1468

*So, you have an overdrawn bank account; we talk about that with negative numbers.*1474

*So, that seems pretty reasonable; we could also talk about √2.*1477

*√2 is able to connect the sides of a square.*1481

*If we have a square where all of the sides are the same on our square, then the connection between one side and the diagonal is side times √2.*1484

*So, we can figure that out from the Pythagorean theorem.*1498

*That makes sense; there is some stuff going on.*1500

*Or if we go ahead and we look at a circle, we will see π showing up.*1502

*If we want to talk about the circumference of a circle (pardon my circle; it is not quite perfect), it will be π times 2 time the radius.*1506

*So, there is π showing up; or, if we wanted to talk about the area, it will be π times the radius squared.*1515

*So, there are relationships going on in circles.*1520

*And circles are real-life things; we see circles in lots of places; we see spheres and other circular objects in lots of places in real life.*1522

*So, it seems reasonable to count √2, π, -1...they are all valid numbers,*1529

*not because we can hold it in our hand, but because we can use it for totally reasonable things.*1534

* So, ultimately, these numbers are "real" (not to say real numbers, but "real" numbers, numbers that we believe in),*1539

*because they have meaning--because they are useful for something.*1545

*A number is valid, not because we can hold it in our hands, but because it is useful and/or interesting.*1550

*That is what makes a number a valid number that we want to work with--because we can either use it in real things,*1557

*or it is really interesting and fascinating--it is telling us cool stuff.*1562

*After all, math is a language; and in language, we can talk about things that aren't just concrete.*1565

*You can talk about things like "cat" and "tree"; but at the same time, you can also express abstract concepts--things like "justice" and "freedom."*1571

*You can walk down the street, and you can point at a cat, and you can point at a tree.*1581

*But you can't really hold a justice in your hand; and you can't say, "Oh, look, here is a freedom."*1585

*They are not things that you can hold; they are not tangible, real things.*1590

*They are abstract concepts that require us to think in this other way.*1594

*And that is how the numbers work: 1, 2, 3...they are representing concrete things that we can really hold.*1598

*But we can also talk about abstract ideas, like √2 or π, that are telling us relationships that are really useful.*1603

*We might not be able to hold it in our hand; but it is still a really useful idea.*1609

*So, it is just as valid; "cat" and "justice" are both valid things, because they are useful to us.*1613

*They represent something worthwhile; they represent something interesting.*1619

*It is the exact same way with the complex numbers; this is how it is with the complex numbers.*1624

*You can't hold i rocks in your hands; you can't hold 52i in your bank account.*1628

*But they still have validity; there is still meaning there; they are still valid; they still have meaning.*1634

*In fact, they have direct connections to the real world; so that might be our other issue:*1641

*"OK, I can believe in the fact that numbers get to be valid when they are interesting; but are they useful--can we use them in the real world?"*1646

*Sure enough, you can: complex numbers show up a lot in electrical engineering.*1653

*They show up in advanced physics; and they show up in other fields of science.*1657

*They also show up in lots of advanced mathematics.*1660

*If you are interested in mathematics--in the really, really high, interesting stuff--complex start to show up a lot.*1663

*They are totally valid; you can prove real things; they are really meaningful.*1667

*By using complex numbers, we can actually model real-world phenomena; and we can make accurate predictions.*1671

*Complex numbers are proven to be useful; we can actually use a complex number and get truth out of it that we can then measure in the real world.*1677

*You don't get a complex number of things; but you can have a complex number help you on your way*1685

*to finding an accurate measurement, to finding something and predicting something that actually works.*1690

*So, complex numbers are totally valid in terms of being useful in the real world, and also just as a thought construct.*1694

*In many ways, the name "imaginary" is unfortunate: they are not imaginary in terms of "they don't count; they aren't really there."*1700

*They are just imaginary because the name stuck; there is no reason that they are less valid than real numbers.*1709

*They aren't less valid; they are just as valid as any other number.*1714

*They are not real numbers, which is to say they are not ℝ; they are not those numbers that we talked about before.*1717

*But the complex numbers can still represent reality.*1723

*So, they are not real numbers, but they still show reality.*1725

*They are imaginary, but only in name; they are actually things that can be used to show real life.*1728

*They tell us all sorts of useful things, and they are pretty cool.*1734

*Complex numbers are legitimate and valid; they are not real numbers, but they are "real" in the sense that they are a part of the real world.*1737

*All that said, nonetheless, complex numbers are not going to be something that we will see a lot.*1746

*They are totally legitimate; they are valid; but we won't see much of them.*1751

*Complex numbers tend to be connected to advanced math, for the most part.*1755

*And so, it is really going to be more advanced math than we want to study right now.*1758

*So, if you keep going in math, or you keep going and see some really high-level science at some point in a few years,*1763

*you will probably end up seeing complex numbers be used for real things.*1768

*But right now, we are just sort of saying, "Oh, look--complex numbers! That is cool," and we are moving on to something else.*1772

*So, most math courses--especially courses at this level--will limit themselves to just the real numbers,*1778

*because if they go too far, it will get too complex (get the joke?).*1783

*Unless a question specifically asks about complex numbers, or they were directly mentioned in the lesson*1791

*(such as this one), just stick to the real numbers.*1796

*You really want to just stick to the real numbers, unless you are working specifically with the complex numbers,*1798

*or you have been told to work specifically with the complex numbers.*1803

*We will briefly play with complex numbers in a couple of lessons in this course.*1805

*But they are something best explored later on in a more advanced mathematics course, or an advanced science course.*1808

*Thus, in general, limit yourself to using just the real numbers, ℝ, for now.*1814

*And really, that is going to be pretty easy, because it is what you are used to doing.*1819

*You are used to just working with the real numbers; so it is not going to be hard to just go back to working with the real numbers,*1823

*because it is what you have been doing for years and years and years.*1827

*All right, we are ready for some examples.*1830

*Simplify (25 - 45i)/(-3 + 4i); remember, we need to multiply by the conjugate.*1832

*The conjugate to -3 + 4i, which we could denote with a bar over all the top of it, is equal to -3...*1838

*and then we flip the sign on the imaginary part, so it will be - 4i.*1844

*So, we want to multiply this by (-3 - 4i)/(-3 - 4i).*1848

*Now, notice: you have to put parentheses around all of this, because the whole thing is multiplying--not just bits and pieces, but the whole thing.*1856

*So, we work this out; 25 times -3 becomes -75; 25 times -4i becomes -100i; -45i times -3 will become positive 45...*1864

*that is 3 times 5 off of 150, or + 135i; -45 times -4 becomes positive, so we will get 4 times 5 off of 200, so 180i.*1878

*Divided by...-3 times -3 gets us positive 9; -3 times 4i gets us + 12i; +4i times -3 gets us -12i; 4i times -4i gets us -4i ^{2}.*1896

*So, we see that we have -12i + 12i, and also when we have i ^{2}, it becomes positive.*1911

*Oops, I accidentally made a typo here: -45 times -4i will become + 180i ^{2}.*1916

*So, cancel out that i ^{2}; we get -180; now, let's combine things.*1923

*-75 - 180; that will get us -255; -100i + 35i will get us +35i; what is on the bottom?*1928

*4 ^{2}...we missed that; sorry--one more mistake; 4 times 4 gets us -4^{2}i^{2},*1938

*so 4 ^{2} gets us 16; 9 + 16 is in our division, so divide by 25.*1949

*-255 + 35i; divide by 25; we notice that we can pull out a 5 from all of these; this is 5 times 51.*1957

*This is 5 times 7; this is 5 times 5; so we go through and cancel one of the 5's on all of them.*1964

*And we are left with -51 + 7i, all over 5, which, if we wanted to, we could alternately represent as -51/5 + 7/5 i,*1972

*keeping our imaginary part and our real part completely separate.*1987

*Both of these are totally legitimate answers; we would know what we were talking about in either case.*1990

*All right, the second example: Given that x = -2 + i is a root to the below polynomial, find the other root and verify both.*1994

*Remember: if x = -2 + i is one of our roots, the conjugate is also the case.*2002

*So, x bar, the conjugate of x being -2 + i, is going to be...what is the conjugate of that?...-2 - i.*2007

*So, we know what the other root is; the other root is -2 - i, and our first root is -2 + i.*2015

*We are guaranteed that a complex conjugate must be the other root, from what we talked about earlier.*2021

*So now we are told to verify both of them.*2027

*There are two different ways we can verify this.*2029

*First, we could verify this through factors; we could show that, if we were to use these as factors...*2031

*because remember, knowing a root tells you a factor; remember, if we know that there is a root at k,*2038

*then we know that there is a factor, (x - k); so if we know that there is a root at (-2 + i),*2046

*then we know that there is a factor of (x - -2 + i), following that same pattern of x - k.*2051

*It is just that k, in this case, is two things.*2058

*That is times (x - (-2 - i)) for our other factor.*2061

*So, if we can multiply these two factors together, and we can get x ^{2} + 4x + 5,*2067

*then we will have verified that those must be the roots, because they are the factors,*2071

*and there is this deep connection between roots and factors; you can go either way.*2075

*So, let's work this out: simplify the insides first: x minus a negative will become + 2 - i;*2078

*times x minus a minus will become + 2 + i; we can start working this out.*2086

*x times x becomes x ^{2}; x times 2 becomes + 2x; x times i will become + ix.*2093

*2 times x will become + 2x; 2 times 2 will become + 4; 2 times i will become + 2i; -i times x will become -ix;*2100

*-i times 2 will become -2i; -i times +i will become -i ^{2}.*2109

*-i ^{2} becomes +1, because the i^{2} cancels out.*2116

*And now, let's work through and see this.*2120

*So, let's simplify this: x ^{2}: how many other x^{2}'s do we have?*2122

*That is the only one, so we get x ^{2} + 2x; how many other x's do we have?*2125

*We have x there, 2x there, and no other x's; so we put those all together, and we get + 4x.*2130

*ix's--how many ix's do we have? We have that ix and that ix, so ix - ix.*2137

*They cancel each other out, and they completely nullify each other; so we don't have to put them down at all.*2142

*How many constants do we have? 4 there; don't forget the 1 that came out of our i ^{2}.*2147

*So, we have 4 + 1, because it flipped the sign; that is + 5.*2152

*And then 2i - 2i; once again, they nullify each other, so we get x ^{2} + 4x + 5; it checks out; great.*2156

*We found the answer.*2163

*The alternate way that we could do this is: we could do this by verifying that they are, indeed, roots.*2164

*So, we could do this another way by showing that they are roots; let's start by showing that x = -2 + i is a root.*2171

*We plug that in; x ^{2}, (-2 + i)^{2}, plus 4(-2 + i), plus 5.*2182

*(-2 + i) ^{2} becomes: -2 times -2 becomes positive 4; -2 on i, plus i on -2, becomes -4i; i on i becomes + i^{2}.*2192

*And then, continue on: plus 4 on -2 becomes + -8; 4 on i becomes +4i; and pull down the 5.*2204

*i ^{2} becomes -1; notice that we have + 4i - 4i, so they eliminate each other here and here.*2213

*4 - 1 becomes 3; -8 + 5 becomes -3; and we get 0...sure enough, that is a root, because it produces 0.*2223

*The other one: let's plug in x = -2 - i; we plug that one in: (-2 - i) ^{2} + 4(-2 - i) + 5.*2232

*-2 times -2 is positive 4; -2 on -i and -i on -2 get us + 4i; -i on -i gets us + i ^{2}.*2244

*Plus -8, minus 4i, plus 5...so we see that we have a positive 4i here and a negative 4i here; they eliminate each other.*2253

*We have this i ^{2}; it becomes -1; so 4 and -1 gets us 3; -8 and 5 gets us -3, which, once again, equals 0; so they are both roots.*2264

*There are two different ways to do it: we can show that these are the factors that would be given by those roots,*2276

*and when you multiply those factors, you get back exactly to where you started; that checks out.*2280

*Or alternately, we can do it by roots and show that when you plug that in, you get the zeroes; so that checks out.*2286

*Great; the third example: Factor x ^{2} - 8x + 19.*2291

*Well, we know that this is probably going to involve complex numbers; it is probably a little bit hard to figure it out in terms of complex numbers.*2297

*But can we find the roots? Sure enough, we can find the roots.*2302

*Let's find roots, and then we will use the roots to give us factors.*2305

*Remember: once you know roots, you know factors; so we find the roots first.*2309

*We can just use the quadratic formula, because now we can use it on anything.*2313

*We don't have to worry about if it is a complex or not.*2317

*The discriminant won't hold us back, because now we can just get imaginary answers, as well.*2319

*We have x =...the roots occur at [-b ± √(b ^{2} - 4ac)]/2a.*2324

*And hopefully, you were able to say that out loud before I said it, to yourself, because really, you want to have that one memorized.*2335

*I said it the last time we talked about the quadratic formula.*2342

*The quadratic formula comes up enough in math and science that it is ultimately something you really want to have memorized.*2344

*All right, so what is our b? Our b is -8.*2349

*So, we plug that in: [-(-8) ± √((-8) ^{2} - 4 (what is our a? our a is a 1) (1) (times...what is our c? c is 19)(19)...*2353

*let's move that square root over all of the way; 2 times...a is 1 again, so 2 times 1.*2367

*That equals -(-8) (gets us positive 8), plus or minus the square root of...64; what is 4 times 19? that is 76, so minus 76; all over 2.*2373

*We divide out the 2, so we will get 8/2; that gets us 4; plus or minus the square root of 64 - 76; that will still be over 2.*2387

*Let's put it over that, just so we don't forget that.*2395

*64 - 76 gets us -12; so we have 4 ± √-12...so we can pull that out as an i, so we will get √12 i, over 12,*2397

*equals 4 ± √12...what is √12? √12 we can see as √4(3), which equals 2√3,*2412

*so plus or minus 2√3 i, over 2; look, we have 2 and 2; those cancel out, and we are left with all of our roots.*2425

*They are when x is equal to 4, plus or minus the square root of 3, times i.*2436

*Those are our roots; however, those aren't our factors.*2442

*We want to find what the factors are; so let's get that in another color.*2446

*If we know that our roots are 4 ± √3i, remember: if you know k is a root, then that tells you x - k is a factor.*2450

*So, in this case, our roots are x = 4 + √3i, and x = 4 - √3i, which is good, because they came as a conjugate pairing there.*2463

*So, those are both of our possibilities; those are both of our factors.*2477

*x - k: our factors will be x minus this one right here, so minus (4 + √3i)...not that whole thing...*2480

*I put that parenthesis on the wrong place; i...the parentheses close there; times (x - this thing here, (4 - √3i).*2494

*So now, let's simplify it, so we can get the factors in a nice, slightly-simpler form to look at.*2506

*x - 4 - √3i and x - 4 + √3i; we have factored it by being able to do that.*2510

*And if we wanted to, we could also expand this and check this.*2522

*And we would be able to show that that is, indeed, exactly what it is; great.*2524

*The final example: What is i ^{3}, i^{4}, i^{5}, i^{6}, i^{7}, i^{8}, etc.?*2528

*What pattern appears as we go through these powers of i?*2535

*Let's take a look at how we work through it.*2538

*If we have i ^{1}, just plain i, we have i.*2541

*That is just what it is; it is just i.*2545

*What about when we have i ^{2}? Well, by definition, that was -1.*2547

*So, let's see the way it keeps going as we take this up.*2551

*i ^{3}...we multiply the -1 by one more i, so we would get -1 times i, or just -i.*2553

*i ^{4} would be equal to...i times i gets us -i^{2}; -i^{2}(i^{2}) cancels, and we get positive 1.*2559

*-i ^{2} cancels, and we get positive 1; so we are left at 1, just a plain + 1.*2567

*What if we keep going? i ^{5} is equal to...well, we multiply by 1, so it is just i, once again.*2578

*i ^{6} would be equal to i^{2}, multiplying by one more i, which we know is -1.*2584

*i ^{7} is equal to i^{3}, which is equal to...we already figured this out; that was -i.*2590

*i ^{8}...well, that is going to be equal to i^{4}, because we just multiply the one above.*2597

*We already figured out what i ^{4} is; that is going to be positive 1.*2602

*Let me make that plus sign a little clearer.*2607

*i ^{9}...if we just kept going, we would have i^{5};*2608

*we already figured out what i ^{5} was--that was i^{1}, which is just i; and so on, and so on, and so on.*2611

*So, the pattern repeats every 4.*2617

*What we need to do is: we basically need to divide by 4 and see what we have.*2627

*What we can do is divide the exponent of i by 4; then, what do we do next?*2632

*Let's do a quick check: if we did i ^{9}, 4 goes into 9 how many times?*2643

*It goes in twice; so we would have 8; 9 - 8 is 1, so we would get a remainder of 1.*2648

*So then, you look at the remainder, and that tells you that it is equal to i to whatever-you-just-figured-out-your-remainder-is.*2653

*So, for example, if we wanted to figure out what i ^{80} is (which is divisible by 4),*2674

*we can see that is just i to the 4 times 4 times 4 times 4 times 4; if we figure that out for i ^{80},*2681

*then we can figure out that what that is equivalent to...by 4...how many times does that go into 80?*2687

*4 goes into 8 twice, so that gets us 8 - 0; bring down the 0; 0; we get 20, and our remainder is 0.*2692

*So, that would be the equivalent of i ^{0}, which is just the same thing as i^{4}, which is +1.*2700

*So, that is how you want to do it if you are given a really, really, really large i.*2708

*It is just a question of if you divided it by 4--what would be left over? What would be the remainder?*2712

*And if you end up having a remainder of 0, then it fit perfectly, so it ends up coming out just as 1.*2716

*All right, great; we will see you at Educator.com later.*2721

*And we will finally see how complex numbers tell us something about polynomials, more than just quadratics.*2723

*We will see how they are deeply connected to everything that we have been talking about.*2728

*It will be so deep that it is called the fundamental theorem of algebra.*2731

*All right, see you later--goodbye!*2735

2 answers

Last reply by: Tiffany Warner

Thu Jun 2, 2016 6:39 PM

Post by Tiffany Warner on June 1, 2016

Hi Professor!

In one of the practice problems, we are asked to find the roots of x^2-6x+20

The answer I continuously got was 3+sqrt(11)i and 3-sqrt(11)i

The answer given does not match. I looked at the steps and there’s two places where I think I’m missing something. Under the square root they show 36-100 but I got 36-80. I’m not sure where the 100 is coming from.

Then in the last step I noticed the first real number is divided by the denominator, but the real number in the “imaginary part” was left alone. Can we do that?

I often make small silly mistakes that lead me to strange conclusions so I’d thought it best to get clarification.

Thank you very much for your time and help!

3 answers

Last reply by: Professor Selhorst-Jones

Mon Feb 16, 2015 12:14 PM

Post by Andre Strohhofer on June 19, 2013

How would you graph these numbers, or how would the roots show up on a graph?