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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (24)

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.

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

Simplify √{ − 50n2m4}
  • Use the Product Property of Square Roots to factor out perfect squares
  • √{ − 50n2m4} = √{ − 1*25*2*n2*m4} = √{ − 1} *√{25} *√2 *√{n2} *√{m4}
  • Simplify and look out for the Principle Square Roots.
  • √{ − 1} *√{25} *√2 *√{n2} *√{m4} = i*5*√2 *|n|*m2
i*5*√2 *|n|*m2
Simplify (4 − 5i) + (10 − 2i) − ( − 4 − 3i)
  • 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
18 − 4i
Simplify (4 − 5i) − (1 − 3i) − ( − 2 − 5i)
  • 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
5 + 3i
Multiply (3 − 2i)(3 + 5i)
  • Multiply using FOIL - First Outter Inner Last
  • 3*3 + 3*5i + ( − 2i)(3) + ( − 2i)(5i)
  • Simplify
  • 9 + 15i − 6i − 10i2 = 9 + 9i − 10i2
  • Recall that by definition i2 = − 1, subsitute i squared
  • 9 + 9i − 10i2 = 9 + 9i − 10( − 1) = 9 + 10 + 9i =
19 + 9i
Multiply (3 + 4i)(4 − 3i)
  • Multiply using FOIL - First Outter Inner Last
  • 3*4 + 3*( − 3i) + 4i*4 + (4i)( − 3i)
  • Simplify
  • 12 − 9i + 16i − 12i2 = 12 + 7i − 12i2
  • Recall that by definition i2 = − 1, subsitute i squared
  • 12 + 7i − 12i2 = 12 + 7i − 12( − 1) = 12 + 12 + 7i =
24 + 7i
On a Complex Plane, plot the following complex numbers a) 5i b) − 2i c)2 + 3id)5 − 3ie)4
  • Recall that complex plane is divided between the Imaginary Part ( y - axis) and Real Part(x - axis).
  • Plot each point on the corresponding location.
On a Complex Plane, plot the following complex numbers a) − i b) − 2 + 2i c) − 5 − 3id) − 1 + 0ie) − 3 − 2i
  • Recall that complex plane is divided between the Imaginary Part ( y - axis) and Real Part(x - axis).
  • Plot each point on the corresponding location.
Simplify [( − 9 − 9i)/3i]
  • 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 i2 = − 1
  • [( − 9 − 9i)/3i]*[3i/3i]
  • Distribute
  • [( − 9 − 9i)/3i]*[3i/3i] = [( − 27i − 27i2)/(9i2)]
  • Subsittue i2 = − 1
  • [( − 27i − 27( − 1))/(9( − 1))] = [(27 − 27i)/( − 9)]
  • Write in standard form
− 3 + 3i
Simplify [(4 + 2i)/(3 + 4i)]
  • 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 i2 = − 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 a2 + b2
  • [(4 + 2i)/(3 + 4i)]*[(3 − 4i)/(3 − 4i)] = [((4 + 2i)(3 − 4i))/(32 + 42)] = [((4 + 2i)(3 − 4i))/25]
  • Multiply in the numerator
  • [((4 + 2i)(3 − 4i))/25] = [(12 − 16i + 6i − 8i2)/25] = [(12 − 10i − 8i2)/25]
  • Eliminate any i2
  • [(12 − 10i − 8i2)/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
[4/5] − [2/5]i
Simplify [( − 4 + 4i)/(3 + 2i)]
  • 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 i2 = − 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 a2 + b2
  • [( − 4 + 4i)/(3 + 2i)]*[(3 − 2i)/(3 − 2i)] = [(( − 4 + 4i)(3 − 2i))/(32 + 22)] = [(( − 4 + 4i)(3 − 2i))/13]
  • Multiply in the numerator
  • [(( − 4 + 4i)(3 − 2i))/13] = [( − 12 + 8i + 12i − 8i2)/13] = [( − 12 + 20i − 8i2)/13]
  • Eliminate any i2
  • [( − 12 + 20i − 8i2)/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]
− [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

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 √32x5,0054

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, x4, 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 x4 (is x2).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 4x2 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, √49x4/25, I know that this is equal to √49x4/√25.0155

This allows me to simplify these into 7x2/5.0178

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: i2 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 i2 equals -1, and that i equals √-1; both of these are very important.0254

So, let's think about an equation such as this: x2 = -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 -b2 equals bi.0351

For the positive real number in this case, it would be 2: b = 2; the square root of -22 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 sign0692

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 i2.1029

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, -20i2: 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 i2 = -1.1062

This helps us to simplify with multiplication, because, since i2 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 i2 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: 2i2--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: -i2.1277

So, this gives me (simplifying) 3; i + 6i is 7i; plus 2i2; over 9; 3i - 3i...that drops out; minus i2.1299

Now, just recall for a second the important concept that i2 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 -12, 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 a2 + b2.1409

So, if I multiply a + bi times a - bi, I am actually going to get a2 + b2.1420

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 a2, which is 32, plus -12, 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

x2 and y4 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 x2; 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 x2 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 y4, actually...the square root of that is y2, 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 y2, 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 y2 will always be positive.1628

Since y2 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|y2.1650

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 -30i2.1802

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 i2 equals -1; so, since i2 = -1, and I have i2 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 i2 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 + 6i2.1971

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 a2 + b2.1994

Now, looking at 4 + 2i here, a equals 4 and b equals 2; so let me just take that shortcut2003

and say that I then have a2 (which is 4, so 42), plus b2 (which is 22).2012

OK, now, simplifying the numerator a bit further gives me 8; -4i - 12i is -16i; plus 6i2.2023

In the denominator, 42 is 16, and 22 is 4.2036

OK, recall that i2 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 a2 + b2."2098

So, that is 42 is 16, and 22 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 + 6i2; and then, I said, "OK, i2 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