WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, I will be introducing the concept of imaginary and complex numbers.
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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.
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Recall the product property: the **product property** and the **quotient property**--both of these properties can be used to simplify square roots.
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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.
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And this allows us to simplify; for example, if you have something like √32x⁵,
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it is not in simplest form, because there are still some perfect squares here under the radical.
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So, the way I can simplify this is to rewrite this as the perfect squares times the other factors.
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Let me rewrite this as the perfect square, which is 16, times 2; so the perfect square of 16 times the other factor, 2.
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And then, for the variable, I have the perfect square, x⁴, times x.
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Now, what this product property allows me to do is then separate out everything like this.
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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).
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And then, I am left with 2x, the square root of 2 times the square root of x.
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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.
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And this is now in simplest form.
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The quotient property is the same idea, but with division.
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If you have something such as √a/b, you can break that down into √a/√b.
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So again, if I have something such as, let's see, √49x⁴/25, I know that this is equal to √49x⁴/√25.
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This allows me to simplify these into 7x²/5.
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So, we are going to be using these properties of square roots in today's (and in future) lessons.
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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.
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And it is not, if you are looking only at the real number system.
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However, there is another number system called the complex number system; and part of that includes imaginary numbers.
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And what these do is: this allows us to find a solution to something like this. What is this?
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Before we go on, let's look up here: what an imaginary number says is that i equals the square root of -1.
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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.
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And this is a concept that is going to become important later on, and we are going to use this.
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So, just recall that i² equals -1, and that i equals √-1; both of these are very important.
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So, let's think about an equation such as this: x² = -4 (related to this).
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We can find a solution to this by saying, "OK, x equals the square root of -4"; previously, we couldn't.
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And the reason is: I can break this out into the following: x = √-1(4).
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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."
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This part is easy to handle: I know that the square root of 4 is 2.
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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."
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So, since the square root of -1 is i, I can simply say that this is i, or x equals 2i.
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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.
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So, that is what this is saying up here.
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For the positive real number b, the square root of -b² equals bi.
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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.
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Looking at this using another example: let's say that I am asked to find x = √-9.
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Again, using the product property, I know that this is equal to -1 times 9.
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And the product property allows me to break this up as such.
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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,
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so I am going to say this is i3, which is usually written as 3i.
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OK, so when you are in a situation where you need to find a negative square root,
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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.
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And the answer is the imaginary number bi.
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Now, I mentioned that imaginary numbers are part of a different number system: we have the real number system,
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and we also have another number system called the complex number system.
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And a **complex number** is in the form a + bi; and there are two parts to this--a real part and an imaginary part.
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So, this is the real part; bi is the imaginary part; and they are complex--they have two parts.
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For example, I could have 5 + 4i; this is a real number; this is an imaginary number; together it is a complex number.
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Or 8 - 3i: again, the real number is 8; the imaginary is -3i.
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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.
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So, this is just a real number; so the real numbers are part of the complex number system.
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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.
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So, when b equals 0, you just end up with a real number.
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Conversely, let a equal 0; then you are going to get 0 + bi.
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This is just a pure imaginary number; this is part of the complex number system, as well--just a pure imaginary number,
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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.
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You can have complex numbers: they have two parts, a real part and an imaginary part.
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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.
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Or you can have both parts and have a complex number.
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OK, equality: this is pretty straightforward--this just says that, if you have a complex number a + bi,
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it equals c + di if and only if the real parts are equal and the imaginary parts are equal.
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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.
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Addition and subtraction: in order to add or subtract complex numbers, combine the real parts and the imaginary parts by addition or subtraction.
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To illustrate this: if you are asked to add 6 + 2i plus 3 + 4i, well, we are told to combine the real parts;
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so I have the real part 6, and I am going to combine that with the real part 3,
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because this is in the form a + bi, where a is real, and then bi is my imaginary part.
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I combine my reals; now the imaginary parts--I have 2i, and I am going to combine that with 4i,
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treating the i like a variable; what you can do is factor it out.
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So, let's pull that i out to give me 2 + 4.
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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.
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So, this is 9 + 6i, just combining the real parts and the imaginary parts.
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Now, working with subtraction: 5 + 2i minus 4 + 6i; OK, it is often simpler to just get rid of this negative sign
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and rewrite this as adding the opposite, like we have done previously with just the real number system.
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This is plus -4, minus 6i; actually, this should be...no, that is correct.
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OK, 5 + 2i minus 4 + 6i: I am rewriting this as 5 + 2i plus -4 - 6i.
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Now, I need to combine the real parts; so the real part I have here is 5 + -4.
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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).
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Now, 5 minus 4 is going to give me 1; plus i--and here I have 2 - 6, so that is -4.
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I am rewriting this as 1 - 4i.
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Again, all I did is changed the signs so that I could just add the opposite of each, plus -4, plus -6i.
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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.
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OK, the complex plane is like the coordinate plane that we have worked with before, but with some important differences.
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So, before, we worked with the coordinate plane, and of course, we had a horizontal and vertical axis.
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And we had positive and negative numbers on it.
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Well, here the horizontal axis represents the real part of a complex number; so this is the real part, or the real numbers.
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On this vertical axis, we have the imaginary numbers--the imaginary part of the complex number.
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For example, this could be labeled 1, 2, 3, 4, 5, and on--pretty familiar.
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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.
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OK, so thinking about graphing complex numbers on a coordinate plane: 2 + 3i, for example:
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well, the 2 is the real part, so that is going to give me my horizontal coordinate right here, 2.
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Now, the vertical coordinate is 3i; so I have 2 here and 1, 2, 3i up here; so this is 2 + 3i.
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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.
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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.
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I may also have a pure imaginary number: maybe I have something like 0 + 2i, so there is no real part to it.
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The real part is just going to be 0; the imaginary part is going to be right here at 2i.
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OK, so to graph a complex number, you find the real part on this horizontal axis, and the imaginary part on the vertical axis.
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Pure real numbers go on the horizontal; pure imaginary numbers go on the vertical axis.
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To multiply complex numbers, you treat them just like any two binomials.
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For example, if you are asked to multiply 3 + 4i and 2 - 5i, I am going to use FOIL;
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and just treat the i's like variables, just as you have in the past, as far as multiplication goes.
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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).
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Now, in a minute, we are going to get into some differences.
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But in these first steps, really, you are just treating it like binomials.
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When you go to simplify, there are differences, though.
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But for the multiplication part, it is familiar territory.
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OK, so working this out: 3 times 2 is 6; 3 times -5i is -15i; plus 4i times 2--that is plus 8i;
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and then, I have 4i times -5i; so this is going to give me -20, and i times i is going to be i².
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Simplify just as you always have: combine like terms.
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I have a 6, and then I have a -15i; and I can combine that with 8i to get -7i.
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OK, -20i²: now, you might think you are done, but you are actually not--
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recall from before that i equals the square root of -1.
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Well, if I square both sides, I mentioned that you would get i² = -1.
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This helps us to simplify with multiplication, because, since i² equals -1,
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I can say 6 - 7i - 20, and I am going to substitute in -1.
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So, I get 6 - 7i; and a negative and a negative is a positive; now, I can simplify...-7i + 26.
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So again, I proceeded with my multiplication, just as I would any two binomials.
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The difference came when I went to simplify, because i² is -1;
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so that actually allowed me to further simplify, because it got rid of that imaginary number.
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I still have an imaginary number here, though, so my result is going to be a complex number, -7i + 26.
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OK, division is a little bit more complicated, but it calls upon some familiar concepts from before.
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With imaginary numbers, we can have what are called complex conjugates.
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Before I go into this, recall the idea of conjugates when we talked about radicals.
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Remember that we said something like √3 + √x has the conjugate √3 - √x.
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And you might recall that we used these conjugate pairs when we were dealing with situations where we had radicals in the denominators.
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We used conjugate pairs, and we would multiply the numerator and the denominator by its conjugate to get rid of radicals in the denominator.
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Here, what I want to do is: now, I want to get rid of complex numbers in the denominator, so that I can divide.
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So, I need to think about conjugates for complex numbers.
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And with complex numbers, it is the same idea: you just reverse the sign before the second term.
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So, if I had a + bi, its conjugate is going to be a - bi.
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For example, if I have 4 + 5i, its conjugate is 4 - 5i.
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To divide complex numbers, multiply both the divisor and the dividend by the conjugate of the divisor.
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In other words, if I have 1 + 2i, 3 - i, here is my dividend; now, what is my conjugate?
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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.
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OK, so I am going to multiply this by 3 + i and this by 3 + i.
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Using my techniques for multiplying two binomials, I am going to get 3, and then Outer terms--that is i;
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the Inner terms--that is going to give me 6i; and then my Last terms: 2i²--using FOIL, just like we always have.
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OK, now in the denominator: 3 times 3 is 9; Outer terms--that is positive 3i; Inner terms: -3i; Last: -i².
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So, this gives me (simplifying) 3; i + 6i is 7i; plus 2i²; over 9; 3i - 3i...that drops out; minus i².
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Now, just recall for a second the important concept that i² equals -1.
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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.
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In the denominator: 9 minus -1 squared; this is going to give me 3 + 7i - 2, over 9 minus -1², which is 1.
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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.
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Again, i squared is equal to -1, so this entire term would just be -1.
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Simplifying 3 - 2 gives me 1 + 7i, over 9 + 1 (is 10).
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So, you see what happened: by multiplying both the numerator and the denominator by the complex conjugate,
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I was able to eliminate the complex number in the denominator.
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Now, just to show you a little bit of a shortcut: when you multiply a complex number by its conjugate, you get a² + b².
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So, if I multiply a + bi times a - bi, I am actually going to get a² + b².
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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.
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If you look at it this way, if I have 3 - i here, a equals 3, and b equals -1.
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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)
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is a², which is 3², plus -1², or 9 + 1, which equals 10.
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And that is a good shortcut to use: I multiplied it out just to show you how this term drops out, and this term--
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you get rid of the imaginary number, and you just end up with the real number down here.
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But it is really a good idea to use shortcuts when you can; it will save you time and mistakes.
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So again, to divide, you multiply both the divisor and the dividend by the conjugate of the divisor.
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OK, in our first example, we are going to use some of the concepts of properties of square roots, and also of complex numbers.
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Using the product property, I can rewrite this so that I can factor out the perfect squares and deal with the imaginary number.
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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.
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x² and y⁴ are also perfect squares.
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The product property tells me that this is equal to this.
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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.
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The square root of 36 is 6; I can't simplify √2 any further.
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Now, let's look at x²; be careful with this, because what they are asking for is the principal, or positive, square root.
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To make this more concrete--how you have to handle this--let's think about if I was told that x² equals 4.
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If I were to take the square root, well, the square roots of that are +2 and -2.
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And the reason is because -2 squared equals 4, and 2 squared equals 4.
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But when I use the radical sign here, what I really want--I am saying I want the principal, or positive, square root.
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So, in order to ensure that I am expressing that, I need to use absolute value bars.
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If I wanted the square root that is a principal square root, I could say it is the absolute value of x.
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And since x equals +2 or -2, the absolute value of x here would just be 2.
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OK, now y⁴, actually...the square root of that is y², and I don't need absolute value bars.
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And let's think about why: I don't know what y is; let's say y stands for -3.
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Well, when I take y², I would get a positive number.
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So, it doesn't matter if y is negative; it doesn't matter if y is positive, because y² will always be positive.
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Since y² is always positive, I don't need to specify absolute value,
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whereas x could be negative, so I do need to specify an absolute value.
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OK, I am just rewriting this as 6i√2|x|y².
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Simplifying this using properties of square roots and the properties of imaginary numbers...knowing that i is √-1 allowed me to simplify this.
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OK, Example 2 involves addition and subtraction of imaginary numbers.
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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.
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I am going to take the opposite of -3, which is 3, and the opposite of -4i, which is positive 4i.
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Now, remember that, to add complex numbers, you add the real parts to each other, and the imaginary parts.
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So, let's look at what I have for real parts: I have 4; I have 7; and I have 3.
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OK, I am adding the real parts; I am combining those; and I am combining the imaginary parts.
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Here, I have -3i; I have 2i; and I have 4i.
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Let's factor out the i, so all I have to do is add these real numbers in here.
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7, 4, and 3 is simply 14; plus i, times -3, 2, and 4; so that is 6 minus 3, which is 3.
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I am rewriting this as 14 + 3i; again, working with complex numbers, adding and subtracting,
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you add the real parts to each other and the imaginary parts to each other.
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In this example, we are multiplying some complex numbers; and you handle these just as you handle any other binomials, using FOIL.
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Multiply out the First terms: that is 4 times 3; the Outer terms: 4 times 6i; the Inner terms--
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this is going to give me + -5i, times 3; and then the Last terms: -5i times 6i.
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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.
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Simplify a bit more to get 12; I can combine these two imaginary numbers; 24i - 15i is positive 9i.
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I can take it one step further with the simplification.
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Recall that i² equals -1; so, since i² = -1, and I have i² right here, I can substitute -1 right here.
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12 + 9i - 30(-1) gives me 12 + 9i...a negative and a negative is a positive, so that gives me + 30.
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Now, I can combine these two: 30 + 12 gives me 42 + 9i.
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Multiply these out just like any two binomials.
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Then, I got to this point; I combined like terms; and right here, I stopped and realized that i² is equal to -1.
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So, I substituted that here, which turned this into a real number that I added to 12; and this is my answer.
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OK, simplify: now we are working with division.
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Remember that, in order to divide, what I need is to multiply the divisor and the dividend by the conjugate of the divisor.
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I am looking here at 4 + 2i; its complex conjugate is 4 - 2i.
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I need to multiply this numerator and denominator by 4 - 2i.
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OK, in the numerator, I am going to go ahead and do my FOIL.
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First gives me 2(4), which is 8; Outer terms--this is 2(-2i)--that is -4i.
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Inner terms: -3i(4) is -12i; Last terms: -3i(-2i) is + 6i².
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Now, in the denominator, I could use FOIL and multiply it out; or I could remember that, if I multiply complex conjugates,
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a + bi times a - bi, what I am going to end up with is a² + b².
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Now, looking at 4 + 2i here, a equals 4 and b equals 2; so let me just take that shortcut
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and say that I then have a² (which is 4, so 4²), plus b² (which is 2²).
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OK, now, simplifying the numerator a bit further gives me 8; -4i - 12i is -16i; plus 6i².
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In the denominator, 4² is 16, and 2² is 4.
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OK, recall that i² is -1; so I am going to substitute -1 here.
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In the denominator, I just have 16 + 4 is 20; this gives me 8 - 16i; this is -6 over 20;
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I can simplify a bit more, because 8 - 6 is 2; this is 2 - 16i, over 20--that is my solution.
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OK, so in order to simplify this, I took the conjugate of the denominator (which is 4 - 2i),
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and I multiplied both the divisor and the dividend by this complex conjugate.
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In the denominator, it was easy, because I just said, "OK, multiplying these conjugates gives me a² + b²."
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So, that is 4² is 16, and 2² is 4, to get 20.
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In the denominator, I used FOIL; I multiplied these out, just as I normally would, to get this.
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I combined like terms to get 8 - 16i + 6i²; and then, I said, "OK, i² is -1,"
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allowing me to simplify this into -6 and combining 8 - 6 to get 2 - 16i over 20.
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That concludes this session of Educator.com introducing complex numbers and imaginary numbers.
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I will see you next time!