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Lecture Comments (7)

2 answers

Last reply by: Tiffany Warner
Thu Jun 2, 2016 6:39 PM

Post by Tiffany Warner on June 1 at 07:44:48 PM

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?

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, x2+1 has no roots and cannot be broken down any further.
  • We introduce a new number to get around this: the imaginary number,



    = i.
    Notice that i2 = −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:
    a   +   b i,
    where a and b are both real numbers. [a is the real part, bi is the imaginary part.]
  • 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 i2 = −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.
    x =
    −b ±


    b2 −4ac



    Previously, we couldn't use the formula if b2 − 4ac < 0, but now we know it just produces an imaginary number! Furthermore, because of the ±√{b2−4ac} 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 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

Transcription: Complex Numbers

Hi--welcome back to

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 x2 + 1 = 0.0016

There are no roots to x2 + 1, because for that to be true, we would need x2 = -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 x2 = -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 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 i2, 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 i2 is, once again, -1; so it checks out--both of these are, indeed, solutions.0174

So, we have found how to solve x2 = -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 on c gets us bci, and bi on di will get us bdi2.0570

Now, remember: i2 = -1; so when we have this i2 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 i2 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 i2 = -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 2i2.0632

Once again, i2 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 30i2.0658

Once again, remember: i2 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 i2 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 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 x2, 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 x2 - 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 - 4i2; so +2i and -2i cancel each other out, because we have the negative here and the plus here.0855

And then, we have 4i2, 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 a2, and we will get - abi + abi;0941

plus here, minus here; those cancel out; i2 is a -1, so that causes that to become a plus.0948

So, we will end up with a2 + b2, 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 10994

(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 c2 + d2; 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 +30i2.1093

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 -4i2.1107

Remember, i2 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, x2 + 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 x2 - ix + ix - i2.1180

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

Those cancel each other out; i2 becomes + 1, so we get x2 + 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 b2 - 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, b2 - 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 ± √(b2 - 4ac) part in the quadratic formula,1236

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

If b2 - 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 root1274

(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 way1685

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 lesson1791

(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 -4i2.1896

So, we see that we have -12i + 12i, and also when we have i2, it becomes positive.1911

Oops, I accidentally made a typo here: -45 times -4i will become + 180i2.1916

So, cancel out that i2; 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

42...we missed that; sorry--one more mistake; 4 times 4 gets us -42i2,1938

so 42 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 x2 + 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 x2; 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 -i2.2109

-i2 becomes +1, because the i2 cancels out.2116

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

So, let's simplify this: x2: how many other x2's do we have?2122

That is the only one, so we get x2 + 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 i2.2147

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 x2 + 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; x2, (-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 + i2.2192

And then, continue on: plus 4 on -2 becomes + -8; 4 on i becomes +4i; and pull down the 5.2204

i2 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 + i2.2244

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

We have this i2; 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 x2 - 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 ± √(b2 - 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 ± √ 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 i3, i4, i5, i6, i7, i8, 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 i1, just plain i, we have i.2541

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

What about when we have i2? Well, by definition, that was -1.2547

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

i3...we multiply the -1 by one more i, so we would get -1 times i, or just -i.2553

i4 would be equal to...i times i gets us -i2; -i2(i2) cancels, and we get positive 1.2559

-i2 cancels, and we get positive 1; so we are left at 1, just a plain + 1.2567

What if we keep going? i5 is equal to...well, we multiply by 1, so it is just i, once again.2578

i6 would be equal to i2, multiplying by one more i, which we know is -1.2584

i7 is equal to i3, which is equal to...we already figured this out; that was -i.2590

i8...well, that is going to be equal to i4, because we just multiply the one above.2597

We already figured out what i4 is; that is going to be positive 1.2602

Let me make that plus sign a little clearer.2607

i9...if we just kept going, we would have i5;2608

we already figured out what i5 was--that was i1, 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 i9, 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 i80 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 i80,2681

then we can figure out that what that is equivalent 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 i0, which is just the same thing as i4, 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 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