Algebra 1 Complex Numbers

Welcome back to www.educator.com.0000

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

I held off on this very special type of number so you could build a lot of things about them that you will see in the section0007

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

Here are some other things that we will cover.0022

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

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

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

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

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

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

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

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

An imaginary number (i) is defined as the √-1.0067

It seems like a small definition and it does not seem to be very useful, especially if we want talk about lots of different numbers.0075

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

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

Suppose I’m looking at the √-7 using some of our properties we can go ahead and break it up into it -1 × 7.0093

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

You will see the definition shows up right there.0105

I'm taking the √ -1.0108

It is that piece that gets turned into an (i) and now I'm representing this number as (i) × √7.0111

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

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

√ -16 would be (i) × 4 or just 4i.0137

How those imaginary numbers relate to our complex numbers?0150

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

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

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

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

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

Maybe I will have a number like 3 – 5i that would be in a nice complex form.0180

I can read off the real and imaginary part.0186

We could throw in some fractions and in here as well, so maybe ½ × 2/3(i).0189

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

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

Notice how we can still say it is in a complex form where the real part is 0.0207

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

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

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

Some of the things that we will have to watch out for is when we get down in the simplification process, so keep a close eye on that.0236

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

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

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

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

Since I'm adding I'm going to go ahead and drop my parentheses here and just figure out which terms I should add together.0269

Notice how all of our real parts I can go ahead and put those together and I can go ahead and collect together the imaginary parts.0280

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

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

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

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

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

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

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

3 – 4 = -1 and 2 – 7(i) = -5(i)0357

This would be written as -1 – 5(i)0369

Again, combine those like terms.0373

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

It looks a lot like the same process.0384

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

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

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

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

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

That reveals that (i)² = -1.0419

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

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

2 – 5(i) multiplied by 1 – 6(i)0432

We will begin by multiplying the first terms 2 × 1 = 2.0436

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

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

minus × minus would be +, 5 × 6 is 30 and now there is my (i)².0465

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

2 – 17(i) now comes this interesting part this (i)² here is the same as -1.0480

We will go ahead and rewrite it as a -1.0491

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

I’m not only multiplying the 30 × -1, but then I can go ahead and combine it with the two out front.0501

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

This one is good.0524

With many of these problems where combining complex numbers you know that you are done0527

when we finally get into that nice complex form.0532

You can easily see the real and imaginary parts.0535

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

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

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

Let us just do some quick example.0563

Let us see if I have 3 + 2(i) its complex conjugate would be 3 – 2(i).0564

If I had -7 – 3(i), the complex conjugate would be -3 + 3(i).0573

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

This will have the feel of rationalizing the denominator that is why we had to cover a lot of that information before working on these complex numbers.0591

I’m going to look at the bottom of this particular division problem 4 + 2(i) ÷ 3 + 5(i).0599

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

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

Multiply on top, multiply on the bottom.0628

This one involves quite a bit of work.0631

I have two terms in my complex number, I will have to foil the top and we will also have to foil out the bottom.0634

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

Let us start with the top.0653

Taking our first terms I have 4 × 3 and that would give me 12.0656

The outside terms would be 4 × -5 (i) = -20(i).0662

On the inside 2(i) + 3(i) then would be 6(i) and our last terms 2 × -5 = -10(i)².0668

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

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

And then we have 5(i) × -5(i) = -25(i)².0695

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

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

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

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

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

Each of these will be need to change into -1.0728

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

I have 12 – 20 + 6(i) = -14(i), -10 × -1 ÷ 9 my outside and inside terms on the bottom cancel -25 × -1.0737

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

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

That one I was talking about how I have the feel of rationalizing the denominator and we are getting rid of those roots in the bottom.0766

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

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

12 – 14(i) + 10 ÷ (9 + 25)0781

12 + 10 = 22 – 14(i) ÷ 340790

It is tempting to try and stop right there but do not do it, we want to continue doing this entire problem0801

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

To get in that form I’m going to use the part where I breakup the 34 and 22 and under the 14.0811

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

This will be 22 ÷ 34 -14(i) ÷ 34.0824

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

I will divide the top and the bottom by 2, 11/17 – 7(i) ÷ 17.0837

This one is actually completely done.0846

I can easily see my real part over here and my imaginary part -7/17.0849

That is I know that this one is done.0855

Division is quite a tricky process remember to multiply by the complex conjugate of the bottom0859

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

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

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

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

In fact the largest part of (i) that we came across was i² and we turned it immediately into a -1.0885

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

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

We see many instances while we are working with (i) and (i) was just equal √1 which of course we call (i).0905

When we have i² and then we thought about squaring the √-1, so we got -1 as its most simplified result.0912

We will see what will happen with i³.0922

One way that you could interpret this would be i² × (i).0927

The reason for doing that is because you can then take the i² sitting right here and simplify that.0934

That will give us -1 × (i) the final result of that would be –i as it is simplest form.0942

Let us choose that same idea and see if we can go ahead and simplify i⁴.0955

That will be i² × i².0959

Both of the i² = -10964

I have -1 × -1 = 10970

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

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

Let us go on to i⁵.0990

This can be thought of as i⁴ × (i).0994

We have already done i⁴, it is way back here, that is equal to just 1.0999

1 × (i) I know this simplifies to (i).1003

On the next one i⁶, i⁴ × i².1010

We have already done i⁴ that is just 1.1020

I will put in i², I will just write across from it -1 so 1 × -1 = -1, very interesting.1024

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

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

Let us see if that continues.1044

I⁷ will be i⁴ × i³, 1 × i³ = -i.1046

Maybe the last one, i⁴ × i⁴, both of those are equal to one, this is equal to 1 as well.1058

Notice how we did starting to see same numbers.1069

There are some good patterns we can see, I mean we got past i⁴ and all of them contained an i⁴1072

so we are getting this similar pieces right here.1079

If I was even to continue on to i⁹ then I will have even more groups of i⁴ and I could see that it would simplified down.1083

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

We can take as much farther and develop a nice shortcut formula so we do not always have to string out into a bunch (i).1104

Let us see how to do that.1111

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

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

To figure out where in that pattern you are at, you can take the exponent and simply divide it by 4 and observe what the remainder is.1129

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

I have made a handy table to figure out what that looks like1142

If I’m looking at (i) to some large power, I will take the power and divide it by 41146

and the remainder happens to be 3, according to my chart here it will simplify to –i.1152

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

Let us take i¹³.1166

If I wanted to I could imagine a whole bunch of i⁴ in here.1173

i⁴, i⁴, i⁴ and then i¹.1178

If we add up all those exponents they add up to 13.1186

You can see that most of them are gone since i⁴ = 1.1191

1 × 1 × 1 × (i) =(i)1196

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

In the shortcut we take the exponent and divide that by 4.1210

13 ÷ 4, 4 goes in the 13 exactly 3 times.1216

Notice why that is important, that is exactly how many bunches of i⁴ I have.1227

When you are going through that division by 4 process you are counting up all of the i⁴ that will simply break down and simplify into 1.1233

It goes in there 3 times, 3 × 4 = 12 and I will subtract that away getting my remainder of 1.1243

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

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

I get the same exact answer as the 4, i¹³ = (i).1267

Now that we know a lot more about these complex numbers let us go through some various examples1275

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

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

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

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

I’m looking at 4 + 5(i) -6 + 3(i)1302

I can see my like terms, let us put together 4 and -6, and 5 and 3.1311

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

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

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

Think foil.1342

Be on the watch out for additional things that we will need to simplify such as the i².1346

Okay, starting off, multiplying our first terms together we will get 1.1351

Our outside terms we will multiply that will be 3(i) then we can take our inside terms 2(i) and then finally take our last terms 2(i) × 3(i) + 6i².1357

We go through and start combining everything we can.1377

1 + I have my like terms here, so 3 + 2 = 5(i) + 6 and take the i² and turn it into -1.1380

Let us continue 1 + 5(i) - 6 now we can see we have just a couple of more things that we can go ahead and combine.1397

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

Let us divide these two complex numbers.1420

I have 2 – 5(i) ÷ 1 – 6(i)1422

This is the lengthy process where we find the complex conjugate of the bottom of if we multiply the top and bottom of our fraction by that.1426

Let us first find that complex conjugate.1434

That will be 1 + 6(i).1437

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

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

Foil top and foil bottom.1452

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

I’m looking at the top, first terms 2 × 1 =2, outside terms would give me 12(i), inside terms -5(i), and last terms -30(i)².1461

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

On the bottom, first terms 1, outside 6i, inside -6i, and our last terms -36i².1483

It is looking much better, looking pretty good.1496

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

Let us see what else we can simplify.1502

We will go ahead and combine these 2i on the top and we will combine these 2i of the bottom1505

and then we will put these i² to make them both -1.1510

2 and then we will have 7(i) - 30 × -1.1518

On the bottom 1 – 36 -1.1527

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

We continue putting things together 2 + 7(i) + 30 since the negative × negative is a positive and 1 + 36.1538

Just a few more things to put together 32 + 7(i) ÷ 37.1553

This point it is tempting to stop it but keep going until you get it into its nice complex form, we can see that real and imaginary part.1562

I’m going to give 37³² and 37⁷.1570

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

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

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

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

In the first one I have i¹⁰⁷.1607

Let us start off by taking that exponent, 107 and we will divide it by 4.1611

This will figure out all the bunches of i⁴ that we have hiding in there.1618

4 of those in the 10, twice and we get 8.1623

Subtract them away I will end up with 27.1629

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

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

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

It simplifies to –i.1660

If you ever forget the way you use that table, you can always build it very quickly by doing just the first few values of (i).1664

(i) = (i), i² = -1, and we have –I and 1.1674

You quickly have built your table.1682

On to the next one i²⁰¹³ something very large.1686

We will grab the exponent and we will divide it by 4.1692

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

Let us put in our 0 placeholder over the 1 and then bring down the 13.1707

4 goes into 13 3 times and we will get 12.1713

This one shows that my remainder is 1.1720

What does this entire thing simplify down to?1725

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

It simplifies down just to (i).1736

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

Keep track of this handy method for simplifying powers of (i) that way you can take those much higher powers1745

and bring them down to something nice and compact.1751

Thank you for watching www.educator.com.1754