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### Imaginary and Complex Numbers

- Imaginary numbers are the square roots of negative numbers.
- All the properties of square roots extend to radicands containing complex numbers.
- The complex numbers satisfy the commutative and associative properties for addition and multiplication.
- To divide one complex number by another one, write the division as a fraction. Then multiply numerator and denominator of this fraction by the complex conjugate of the denominator.

### Imaginary and Complex Numbers

^{2}m

^{4}}

- Use the Product Property of Square Roots to factor out perfect squares
- √{ − 50n
^{2}m^{4}} = √{ − 1*25*2*n^{2}*m^{4}} = √{ − 1} *√{25} *√2 *√{n^{2}} *√{m^{4}} - Simplify and look out for the Principle Square Roots.
- √{ − 1} *√{25} *√2 *√{n
^{2}} *√{m^{4}} = i*5*√2 *|n|*m^{2}

^{2}

- Distribute the negative
- (4 − 5i) + (10 − 2i) − ( − 4 − 3i) = (4 − 5i) + (10 − 2i) + (4 + 3i)
- Add the Real parts and add the Imaginary Parts together
- (4 + 10 + 4) + ( − 5i + − 2i + 3i) = 18 − 4i

- Distribute the negative sign
- (4 − 5i) − (1 − 3i) − ( − 2 − 5i) = (4 − 5i) + ( − 1 + 3i) + (2 + 5i)
- Add the Real parts and add the Imaginary Parts together
- (4 − 1 + 2) + ( − 5i + 3i + 5i) = 5 + 3i

- Multiply using FOIL - First Outter Inner Last
- 3*3 + 3*5i + ( − 2i)(3) + ( − 2i)(5i)
- Simplify
- 9 + 15i − 6i − 10i
^{2}= 9 + 9i − 10i^{2} - Recall that by definition i
^{2}= − 1, subsitute i squared - 9 + 9i − 10i
^{2}= 9 + 9i − 10( − 1) = 9 + 10 + 9i =

- Multiply using FOIL - First Outter Inner Last
- 3*4 + 3*( − 3i) + 4i*4 + (4i)( − 3i)
- Simplify
- 12 − 9i + 16i − 12i
^{2}= 12 + 7i − 12i^{2} - Recall that by definition i
^{2}= − 1, subsitute i squared - 12 + 7i − 12i
^{2}= 12 + 7i − 12( − 1) = 12 + 12 + 7i =

- Recall that complex plane is divided between the Imaginary Part ( y - axis) and Real Part(x - axis).
- Plot each point on the corresponding location.

- Recall that complex plane is divided between the Imaginary Part ( y - axis) and Real Part(x - axis).
- Plot each point on the corresponding location.

- This is a Complex Number division problem. When dividing complex numbers your goal is to
- leave your answer in standard form (a + bi) where a is the real part and b your imaginary part.
- This is a special case in which the divisor is not a binomial. What you do in this case is mutiply
- numerator and denominator by the divisor. Remember that i
^{2}= − 1 - [( − 9 − 9i)/3i]*[3i/3i]
- Distribute
- [( − 9 − 9i)/3i]*[3i/3i] = [( − 27i − 27i
^{2})/(9i^{2})] - Subsittue i
^{2}= − 1 - [( − 27i − 27( − 1))/(9( − 1))] = [(27 − 27i)/( − 9)]
- Write in standard form

- This is a Complex Number division problem. When dividing complex numbers your goal is to
- leave your answer in standard form (a + bi) where a is the real part and b your imaginary part.
- This is the case in which the divisor is a binomial. What you do in this case is mutiply
- numerator and denominator by the complex conjugate of the divisor. Remember that i
^{2}= − 1 - Multiply by complex conjugate of divisor
- [(4 + 2i)/(3 + 4i)]*[(3 − 4i)/(3 − 4i)]
- Recall that when multiplying a complex number by its conjugate you can use the short - cut a
^{2}+ b^{2} - [(4 + 2i)/(3 + 4i)]*[(3 − 4i)/(3 − 4i)] = [((4 + 2i)(3 − 4i))/(3
^{2}+ 4^{2})] = [((4 + 2i)(3 − 4i))/25] - Multiply in the numerator
- [((4 + 2i)(3 − 4i))/25] = [(12 − 16i + 6i − 8i
^{2})/25] = [(12 − 10i − 8i^{2})/25] - Eliminate any i
^{2} - [(12 − 10i − 8i
^{2})/25] = [(12 − 10i − 8( − 1))/25] = [(12 + 8 − 10i)/25] = [(20 − 10i)/25] - write into Standard form a + bi
- [(20 − 10i)/25] = [20/25] − [10i/25] = [4/5] − [2/5]i

- This is a Complex Number division problem. When dividing complex numbers your goal is to
- leave your answer in standard form (a + bi) where a is the real part and b your imaginary part.
- This is the case in which the divisor is a binomial. What you do in this case is mutiply
- numerator and denominator by the complex conjugate of the divisor. Remember that i
^{2}= − 1 - Multiply by the complex conjugate of the divisor
- [( − 4 + 4i)/(3 + 2i)]*[(3 − 2i)/(3 − 2i)]
- Recall that when multiplying a complex number by its conjugate you can use the short - cut a
^{2}+ b^{2} - [( − 4 + 4i)/(3 + 2i)]*[(3 − 2i)/(3 − 2i)] = [(( − 4 + 4i)(3 − 2i))/(3
^{2}+ 2^{2})] = [(( − 4 + 4i)(3 − 2i))/13] - Multiply in the numerator
- [(( − 4 + 4i)(3 − 2i))/13] = [( − 12 + 8i + 12i − 8i
^{2})/13] = [( − 12 + 20i − 8i^{2})/13] - Eliminate any i
^{2} - [( − 12 + 20i − 8i
^{2})/13] = [( − 12 + 20i − 8( − 1))/13] = [( − 12 + 8 + 20i)/13] = [( − 4 + 20i)/13] - Write into Standard form a + bi
- [( − 4 + 20i)/13] = − [4/13] + [20i/13]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Imaginary and 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
- Properties of Square Roots 0:10
- Product Property
- Example: Product Property
- Quotient Property
- Example: Quotient Property
- Imaginary Numbers 3:12
- Imaginary i
- Examples: Imaginary Number
- Complex Numbers 7:23
- Real Part and Imaginary Part
- Examples: Complex Numbers
- Equality 9:37
- Example: Equal Complex Numbers
- Addition and Subtraction 10:12
- Examples: Adding Complex Numbers
- Complex Plane 13:32
- Horizontal Axis (Real)
- Vertical Axis (Imaginary)
- Example: Labeling
- Multiplication 15:57
- Example: FOIL Method
- Division 18:37
- Complex Conjugates
- Conjugate Pairs
- Example: Dividing Complex Numbers
- Example 1: Simplify Complex Number 24:50
- Example 2: Simplify Complex Number 27:56
- Example 3: Multiply Complex Numbers 29:27
- Example 3: Dividing Complex Numbers 31:48

### Algebra 2

### Transcription: Imaginary and Complex Numbers

*Welcome to Educator.com.*0000

*Today, I will be introducing the concept of imaginary and complex numbers.*0002

*The concept of imaginary numbers is tied in with square roots; so we are going to begin by a review of some of the properties of square roots.*0010

*Recall the product property: the product property and the quotient property--both of these properties can be used to simplify square roots.*0021

*For numbers a and b that are greater than 0, if you have something like this, √ab, that can be broken down into √a, times √b.*0039

*And this allows us to simplify; for example, if you have something like √32x ^{5},*0054

*it is not in simplest form, because there are still some perfect squares here under the radical.*0062

*So, the way I can simplify this is to rewrite this as the perfect squares times the other factors.*0068

*Let me rewrite this as the perfect square, which is 16, times 2; so the perfect square of 16 times the other factor, 2.*0076

*And then, for the variable, I have the perfect square, x ^{4}, times x.*0085

*Now, what this product property allows me to do is then separate out everything like this.*0093

*Once I have that, I can take the square roots of the perfect squares: I have the square root of 16 (is 4); the square root of x ^{4} (is x^{2}).*0108

*And then, I am left with 2x, the square root of 2 times the square root of x.*0118

*And by going the other way with the product property, I can put those back together and say what I end up with is 4x ^{2} times √2x.*0126

*And this is now in simplest form.*0134

*The quotient property is the same idea, but with division.*0137

*If you have something such as √a/b, you can break that down into √a/√b.*0147

*So again, if I have something such as, let's see, √49x ^{4}/25, I know that this is equal to √49x^{4}/√25.*0155

*This allows me to simplify these into 7x ^{2}/5.*0178

*So, we are going to be using these properties of square roots in today's (and in future) lessons.*0185

*OK, a new concept--imaginary numbers: in Algebra I, when we came across something such as √-4, we simply said that it wasn't defined.*0192

*And it is not, if you are looking only at the real number system.*0206

*However, there is another number system called the complex number system; and part of that includes imaginary numbers.*0209

*And what these do is: this allows us to find a solution to something like this. What is this?*0216

*Before we go on, let's look up here: what an imaginary number says is that i equals the square root of -1.*0230

*Looking at that another way: if i equals the square root of negative 1, let's say I square both of these: i ^{2} is -1.*0238

*And this is a concept that is going to become important later on, and we are going to use this.*0249

*So, just recall that i ^{2} equals -1, and that i equals √-1; both of these are very important.*0254

*So, let's think about an equation such as this: x ^{2} = -4 (related to this).*0262

*We can find a solution to this by saying, "OK, x equals the square root of -4"; previously, we couldn't.*0275

*And the reason is: I can break this out into the following: x = √-1(4).*0286

*Using the product property that we just learned, I can say, "OK, x then equals the square root of -1 times the square root of 4."*0298

*This part is easy to handle: I know that the square root of 4 is 2.*0313

*Now, I have a way of handling this: I go back, and I say, "OK, the square root of -1 is i; I have something to define this as."*0317

*So, since the square root of -1 is i, I can simply say that this is i, or x equals 2i.*0326

*Whereas before I couldn't solve this, now I can, because I can pull out this -1 using the product property, and define that as the imaginary number i.*0335

*So, that is what this is saying up here.*0348

*For the positive real number b, the square root of -b ^{2} equals bi.*0351

*For the positive real number in this case, it would be 2: b = 2; the square root of -2 ^{2} is bi; and that gives me this.*0364

*Looking at this using another example: let's say that I am asked to find x = √-9.*0383

*Again, using the product property, I know that this is equal to -1 times 9.*0393

*And the product property allows me to break this up as such.*0403

*This is easy; I know that the square root of 9 is 3; now I have a way to define this as i; √-1 is i,*0408

*so I am going to say this is i3, which is usually written as 3i.*0417

*OK, so when you are in a situation where you need to find a negative square root,*0425

*the thing is to use the product property to pull this square root of -1 out, and then you just need to find the positive square root.*0429

*And the answer is the imaginary number bi.*0436

*Now, I mentioned that imaginary numbers are part of a different number system: we have the real number system,*0444

*and we also have another number system called the complex number system.*0449

*And a complex number is in the form a + bi; and there are two parts to this--a real part and an imaginary part.*0453

*So, this is the real part; bi is the imaginary part; and they are complex--they have two parts.*0464

*For example, I could have 5 + 4i; this is a real number; this is an imaginary number; together it is a complex number.*0475

*Or 8 - 3i: again, the real number is 8; the imaginary is -3i.*0485

*Now, looking a little deeper, let's say I have b = 0; then, what I am going to end up with is a + 0i, which is just a.*0494

*So, this is just a real number; so the real numbers are part of the complex number system.*0509

*For example, if I were to give you 8 + 0i, well, this is going to drop out; and this is just 8, and it is a real number.*0515

*So, when b equals 0, you just end up with a real number.*0526

*Conversely, let a equal 0; then you are going to get 0 + bi.*0529

*This is just a pure imaginary number; this is part of the complex number system, as well--just a pure imaginary number,*0535

*such as (if a is 0) 0 + 4i; 0 drops out, and I just have 4i; and this is just the imaginary part, the imaginary number.*0544

*You can have complex numbers: they have two parts, a real part and an imaginary part.*0559

*If, in the imaginary part, b equals 0, that leaves you with a real number; if a equals 0, it leaves you with a purely imaginary number.*0564

*Or you can have both parts and have a complex number.*0572

*OK, equality: this is pretty straightforward--this just says that, if you have a complex number a + bi,*0577

*it equals c + di if and only if the real parts are equal and the imaginary parts are equal.*0583

*For example, 4 + 7i equals 4 + 7i, because the real parts are equal (4 = 4, so a = a) and the imaginary parts are equal (b = b; 7 = 7)--straightforward.*0593

*Addition and subtraction: in order to add or subtract complex numbers, combine the real parts and the imaginary parts by addition or subtraction.*0613

*To illustrate this: if you are asked to add 6 + 2i plus 3 + 4i, well, we are told to combine the real parts;*0625

*so I have the real part 6, and I am going to combine that with the real part 3,*0637

*because this is in the form a + bi, where a is real, and then bi is my imaginary part.*0643

*I combine my reals; now the imaginary parts--I have 2i, and I am going to combine that with 4i,*0653

*treating the i like a variable; what you can do is factor it out.*0663

*So, let's pull that i out to give me 2 + 4.*0667

*OK, so this leaves me with 6 + 3 (is 9), plus i times 2 + 4 (is 6), but conventionally, we write it with the number first, and then the i.*0672

*So, this is 9 + 6i, just combining the real parts and the imaginary parts.*0686

*Now, working with subtraction: 5 + 2i minus 4 + 6i; OK, it is often simpler to just get rid of this negative sign*0692

*and rewrite this as adding the opposite, like we have done previously with just the real number system.*0709

*This is plus -4, minus 6i; actually, this should be...no, that is correct.*0715

*OK, 5 + 2i minus 4 + 6i: I am rewriting this as 5 + 2i plus -4 - 6i.*0728

*Now, I need to combine the real parts; so the real part I have here is 5 + -4.*0740

*OK, combining the imaginary parts: I have 2i - 6i; this gives me 5 - 4, plus (I want to factor out the i) i times (2 - 6).*0748

*Now, 5 minus 4 is going to give me 1; plus i--and here I have 2 - 6, so that is -4.*0772

*I am rewriting this as 1 - 4i.*0785

*Again, all I did is changed the signs so that I could just add the opposite of each, plus -4, plus -6i.*0788

*Then, I combined the real parts, which were 5 and -4, to give me 1, and the imaginary parts, which were 2i and -6i, to give me -4i: 1 - 4i.*0797

*OK, the complex plane is like the coordinate plane that we have worked with before, but with some important differences.*0810

*So, before, we worked with the coordinate plane, and of course, we had a horizontal and vertical axis.*0819

*And we had positive and negative numbers on it.*0824

*Well, here the horizontal axis represents the real part of a complex number; so this is the real part, or the real numbers.*0826

*On this vertical axis, we have the imaginary numbers--the imaginary part of the complex number.*0838

*For example, this could be labeled 1, 2, 3, 4, 5, and on--pretty familiar.*0850

*What is different here is the vertical axis: here I am going to have i, 2i, 3i, 4i, and the same on down...-i, -2i, -3i, -4i.*0862

*OK, so thinking about graphing complex numbers on a coordinate plane: 2 + 3i, for example:*0879

*well, the 2 is the real part, so that is going to give me my horizontal coordinate right here, 2.*0887

*Now, the vertical coordinate is 3i; so I have 2 here and 1, 2, 3i up here; so this is 2 + 3i.*0896

*Now, imagine I have something that is just a real number, like 5 + 0i; this is going to drop out, so it is just 5.*0907

*Therefore, it is going to be right on the x-axis; I am going to have 5, and then for the vertical plane it is just 0; so, this is 5 + 0i.*0916

*I may also have a pure imaginary number: maybe I have something like 0 + 2i, so there is no real part to it.*0924

*The real part is just going to be 0; the imaginary part is going to be right here at 2i.*0932

*OK, so to graph a complex number, you find the real part on this horizontal axis, and the imaginary part on the vertical axis.*0938

*Pure real numbers go on the horizontal; pure imaginary numbers go on the vertical axis.*0948

*To multiply complex numbers, you treat them just like any two binomials.*0957

*For example, if you are asked to multiply 3 + 4i and 2 - 5i, I am going to use FOIL;*0962

*and just treat the i's like variables, just as you have in the past, as far as multiplication goes.*0970

*Now, I multiply out 3 times 2 (First); then my Outers (and that is 3 times -5i); and then the Inners (+ 4i times 2), and then Last (that is 4i times -5i).*0977

*Now, in a minute, we are going to get into some differences.*1007

*But in these first steps, really, you are just treating it like binomials.*1009

*When you go to simplify, there are differences, though.*1012

*But for the multiplication part, it is familiar territory.*1015

*OK, so working this out: 3 times 2 is 6; 3 times -5i is -15i; plus 4i times 2--that is plus 8i;*1018

*and then, I have 4i times -5i; so this is going to give me -20, and i times i is going to be i ^{2}.*1029

*Simplify just as you always have: combine like terms.*1040

*I have a 6, and then I have a -15i; and I can combine that with 8i to get -7i.*1044

*OK, -20i ^{2}: now, you might think you are done, but you are actually not--*1051

*recall from before that i equals the square root of -1.*1056

*Well, if I square both sides, I mentioned that you would get i ^{2} = -1.*1062

*This helps us to simplify with multiplication, because, since i ^{2} equals -1,*1068

*I can say 6 - 7i - 20, and I am going to substitute in -1.*1074

*So, I get 6 - 7i; and a negative and a negative is a positive; now, I can simplify...-7i + 26.*1082

*So again, I proceeded with my multiplication, just as I would any two binomials.*1095

*The difference came when I went to simplify, because i ^{2} is -1;*1099

*so that actually allowed me to further simplify, because it got rid of that imaginary number.*1104

*I still have an imaginary number here, though, so my result is going to be a complex number, -7i + 26.*1109

*OK, division is a little bit more complicated, but it calls upon some familiar concepts from before.*1117

*With imaginary numbers, we can have what are called complex conjugates.*1125

*Before I go into this, recall the idea of conjugates when we talked about radicals.*1130

*Remember that we said something like √3 + √x has the conjugate √3 - √x.*1135

*And you might recall that we used these conjugate pairs when we were dealing with situations where we had radicals in the denominators.*1147

*We used conjugate pairs, and we would multiply the numerator and the denominator by its conjugate to get rid of radicals in the denominator.*1157

*Here, what I want to do is: now, I want to get rid of complex numbers in the denominator, so that I can divide.*1167

*So, I need to think about conjugates for complex numbers.*1180

*And with complex numbers, it is the same idea: you just reverse the sign before the second term.*1184

*So, if I had a + bi, its conjugate is going to be a - bi.*1190

*For example, if I have 4 + 5i, its conjugate is 4 - 5i.*1200

*To divide complex numbers, multiply both the divisor and the dividend by the conjugate of the divisor.*1207

*In other words, if I have 1 + 2i, 3 - i, here is my dividend; now, what is my conjugate?*1215

*I have 3 - i; the conjugate is going to be 3 + i; so I need to multiply both the divisor and the dividend by 3 + i.*1227

*OK, so I am going to multiply this by 3 + i and this by 3 + i.*1242

*Using my techniques for multiplying two binomials, I am going to get 3, and then Outer terms--that is i;*1252

*the Inner terms--that is going to give me 6i; and then my Last terms: 2i ^{2}--using FOIL, just like we always have.*1267

*OK, now in the denominator: 3 times 3 is 9; Outer terms--that is positive 3i; Inner terms: -3i; Last: -i ^{2}.*1277

*So, this gives me (simplifying) 3; i + 6i is 7i; plus 2i ^{2}; over 9; 3i - 3i...that drops out; minus i^{2}.*1299

*Now, just recall for a second the important concept that i ^{2} equals -1.*1316

*Coming down here, this is going to give me 3 + 7i + 2; and I am going to substitute in -1 here and -1 there; that is going to give me -1.*1326

*In the denominator: 9 minus -1 squared; this is going to give me 3 + 7i - 2, over 9 minus -1 ^{2}, which is 1.*1339

*Correction: this is not squared--this is simply -1: 9 minus -1, so a negative and a negative is going to give me a positive.*1365

*Again, i squared is equal to -1, so this entire term would just be -1.*1378

*Simplifying 3 - 2 gives me 1 + 7i, over 9 + 1 (is 10).*1385

*So, you see what happened: by multiplying both the numerator and the denominator by the complex conjugate,*1394

*I was able to eliminate the complex number in the denominator.*1402

*Now, just to show you a little bit of a shortcut: when you multiply a complex number by its conjugate, you get a ^{2} + b^{2}.*1409

*So, if I multiply a + bi times a - bi, I am actually going to get a ^{2} + b^{2}.*1420

*And that would have allowed me to save a lot of work and a possible mistake down here, because the more work, the more chance of a mistake.*1430

*If you look at it this way, if I have 3 - i here, a equals 3, and b equals -1.*1436

*So, I could just say that what I am going to end up with, if I multiply 3 - i times 3 + i (the complex conjugates)*1447

*is a ^{2}, which is 3^{2}, plus -1^{2}, or 9 + 1, which equals 10.*1453

*And that is a good shortcut to use: I multiplied it out just to show you how this term drops out, and this term--*1461

*you get rid of the imaginary number, and you just end up with the real number down here.*1467

*But it is really a good idea to use shortcuts when you can; it will save you time and mistakes.*1472

*So again, to divide, you multiply both the divisor and the dividend by the conjugate of the divisor.*1477

*OK, in our first example, we are going to use some of the concepts of properties of square roots, and also of complex numbers.*1489

*Using the product property, I can rewrite this so that I can factor out the perfect squares and deal with the imaginary number.*1498

*I have a -1, times 36, times 2; so this is factoring out this -72, so that I have factored out the negative part, and I have factored out the perfect square.*1508

*x ^{2} and y^{4} are also perfect squares.*1523

*The product property tells me that this is equal to this.*1527

*This allows me to simplify: recall that i equals the square root of -1, so instead of writing this, I am going to write it as i.*1537

*The square root of 36 is 6; I can't simplify √2 any further.*1546

*Now, let's look at x ^{2}; be careful with this, because what they are asking for is the principal, or positive, square root.*1552

*To make this more concrete--how you have to handle this--let's think about if I was told that x ^{2} equals 4.*1562

*If I were to take the square root, well, the square roots of that are +2 and -2.*1569

*And the reason is because -2 squared equals 4, and 2 squared equals 4.*1577

*But when I use the radical sign here, what I really want--I am saying I want the principal, or positive, square root.*1584

*So, in order to ensure that I am expressing that, I need to use absolute value bars.*1590

*If I wanted the square root that is a principal square root, I could say it is the absolute value of x.*1596

*And since x equals +2 or -2, the absolute value of x here would just be 2.*1601

*OK, now y ^{4}, actually...the square root of that is y^{2}, and I don't need absolute value bars.*1606

*And let's think about why: I don't know what y is; let's say y stands for -3.*1614

*Well, when I take y ^{2}, I would get a positive number.*1621

*So, it doesn't matter if y is negative; it doesn't matter if y is positive, because y ^{2} will always be positive.*1628

*Since y ^{2} is always positive, I don't need to specify absolute value,*1639

*whereas x could be negative, so I do need to specify an absolute value.*1644

*OK, I am just rewriting this as 6i√2|x|y ^{2}.*1650

*Simplifying this using properties of square roots and the properties of imaginary numbers...knowing that i is √-1 allowed me to simplify this.*1659

*OK, Example 2 involves addition and subtraction of imaginary numbers.*1674

*Simplifying: the first step, to keep my signs straight, is going to be to change this to addition, thus pushing this negative sign inside the parentheses.*1682

*I am going to take the opposite of -3, which is 3, and the opposite of -4i, which is positive 4i.*1694

*Now, remember that, to add complex numbers, you add the real parts to each other, and the imaginary parts.*1701

*So, let's look at what I have for real parts: I have 4; I have 7; and I have 3.*1710

*OK, I am adding the real parts; I am combining those; and I am combining the imaginary parts.*1717

*Here, I have -3i; I have 2i; and I have 4i.*1722

*Let's factor out the i, so all I have to do is add these real numbers in here.*1733

*7, 4, and 3 is simply 14; plus i, times -3, 2, and 4; so that is 6 minus 3, which is 3.*1741

*I am rewriting this as 14 + 3i; again, working with complex numbers, adding and subtracting,*1751

*you add the real parts to each other and the imaginary parts to each other.*1762

*In this example, we are multiplying some complex numbers; and you handle these just as you handle any other binomials, using FOIL.*1769

*Multiply out the First terms: that is 4 times 3; the Outer terms: 4 times 6i; the Inner terms--*1778

*this is going to give me + -5i, times 3; and then the Last terms: -5i times 6i.*1787

*Finish our multiplication, and then simplify: 4 times 3--that is 12, plus 24i; -5i times 3 is -15i; -5i times 6i is going to give me -30i ^{2}.*1802

*Simplify a bit more to get 12; I can combine these two imaginary numbers; 24i - 15i is positive 9i.*1824

*I can take it one step further with the simplification.*1834

*Recall that i ^{2} equals -1; so, since i^{2} = -1, and I have i^{2} right here, I can substitute -1 right here.*1837

*12 + 9i - 30(-1) gives me 12 + 9i...a negative and a negative is a positive, so that gives me + 30.*1859

*Now, I can combine these two: 30 + 12 gives me 42 + 9i.*1877

*Multiply these out just like any two binomials.*1885

*Then, I got to this point; I combined like terms; and right here, I stopped and realized that i ^{2} is equal to -1.*1888

*So, I substituted that here, which turned this into a real number that I added to 12; and this is my answer.*1898

*OK, simplify: now we are working with division.*1909

*Remember that, in order to divide, what I need is to multiply the divisor and the dividend by the conjugate of the divisor.*1912

*I am looking here at 4 + 2i; its complex conjugate is 4 - 2i.*1932

*I need to multiply this numerator and denominator by 4 - 2i.*1941

*OK, in the numerator, I am going to go ahead and do my FOIL.*1953

*First gives me 2(4), which is 8; Outer terms--this is 2(-2i)--that is -4i.*1958

*Inner terms: -3i(4) is -12i; Last terms: -3i(-2i) is + 6i ^{2}.*1971

*Now, in the denominator, I could use FOIL and multiply it out; or I could remember that, if I multiply complex conjugates,*1984

*a + bi times a - bi, what I am going to end up with is a ^{2} + b^{2}.*1994

*Now, looking at 4 + 2i here, a equals 4 and b equals 2; so let me just take that shortcut*2003

*and say that I then have a ^{2} (which is 4, so 4^{2}), plus b^{2} (which is 2^{2}).*2012

*OK, now, simplifying the numerator a bit further gives me 8; -4i - 12i is -16i; plus 6i ^{2}.*2023

*In the denominator, 4 ^{2} is 16, and 2^{2} is 4.*2036

*OK, recall that i ^{2} is -1; so I am going to substitute -1 here.*2043

*In the denominator, I just have 16 + 4 is 20; this gives me 8 - 16i; this is -6 over 20;*2059

*I can simplify a bit more, because 8 - 6 is 2; this is 2 - 16i, over 20--that is my solution.*2074

*OK, so in order to simplify this, I took the conjugate of the denominator (which is 4 - 2i),*2083

*and I multiplied both the divisor and the dividend by this complex conjugate.*2089

*In the denominator, it was easy, because I just said, "OK, multiplying these conjugates gives me a ^{2} + b^{2}."*2098

*So, that is 4 ^{2} is 16, and 2^{2} is 4, to get 20.*2107

*In the denominator, I used FOIL; I multiplied these out, just as I normally would, to get this.*2112

*I combined like terms to get 8 - 16i + 6i ^{2}; and then, I said, "OK, i^{2} is -1,"*2121

*allowing me to simplify this into -6 and combining 8 - 6 to get 2 - 16i over 20.*2128

*That concludes this session of Educator.com introducing complex numbers and imaginary numbers.*2138

*I will see you next time!*2145

1 answer

Last reply by: Dr Carleen Eaton

Sat Nov 1, 2014 4:14 PM

Post by Larry Oldham III on October 21, 2014

How would you solve i^2 + i^3

1 answer

Last reply by: Dr Carleen Eaton

Fri Oct 10, 2014 12:10 AM

Post by Sangeeta Chaudhari on September 27, 2014

Hello Dr. Carleen! I loved your lecture but I don't understand the absolute value part in example 1:Simplify Complex numbers. Can you please explain that because I watch your other videos and I just don't know when to use absolute values. Thanks!

1 answer

Last reply by: Dr Carleen Eaton

Thu Jul 31, 2014 6:52 PM

Post by Prashanti Kodali on July 16, 2014

Dr. Eaton I was doing practice problems and I am unsure if I got this question correct. The question was 3i(2i+4i-6). The answer I got was -12-24i. Can you tell me if I got this question right? Thank you so much!

1 answer

Last reply by: Dr Carleen Eaton

Sun Oct 20, 2013 11:26 AM

Post by dayan assaf on October 19, 2013

so if the question was 1+i, would the imaginary part be 0 or 1?

0 answers

Post by Juan Herrera on September 19, 2013

Difference of two squares:

a^2-b^2 = (a+b)(a-b)

1 answer

Last reply by: Dr Carleen Eaton

Sat Sep 14, 2013 2:54 PM

Post by Chateau Siqueira on September 3, 2013

Where can a find a Lecture about " Difference Quotient" ? Thanks

1 answer

Last reply by: Dr Carleen Eaton

Sat Sep 14, 2013 2:52 PM

Post by Tami Cummins on August 13, 2013

Dr. Eaton in the last example problem when using the short cut and squaring the "b" term which was 2 you didn't include the i's as part of the b term but in the previous sample division problem you did. In the previous sample problem could you say that i had a coefficient of 1 and squared that instead just for consistency?

1 answer

Last reply by: Dr Carleen Eaton

Sat Jul 27, 2013 10:11 AM

Post by Aleksander Rinaldo on July 10, 2013

Under divison, you used (a+bi)(a-bi)=a^2+bi^2....... Why would it not be a^2-bi^2....

0 answers

Post by Victor Castillo on January 24, 2013

Who decided we need imaginary numbers Peter Pan?

0 answers

Post by Victor Castillo on January 24, 2013

Whaaaaat?

2 answers

Last reply by: Norman Cervantes

Mon Apr 29, 2013 12:24 PM

Post by Daniel Cuellar on October 17, 2012

why do you not simplify your answer at the end by dividing everything by 2? just as you would a normal fraction??? please explain.

2 answers

Last reply by: Dr Carleen Eaton

Tue Jul 3, 2012 7:20 PM

Post by David Burgoon on June 27, 2012

Is it necessary to factor out the i in the equations? It looks like you could just add/subtract them. If you could provide an example that supports why it is important, that would help.