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College Algebra: Dividing Complex Numbers

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About this Lesson

  • Type: Video Tutorial
  • Length: 9:07
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 98 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
Trigonometry: Full Course (152 lessons, $148.50)
College Algebra: Basics & Prerequisites (37 lessons, $52.47)
Trigonometry: Complex Numbers & Polar Coordinates (15 lessons, $26.73)
Trigonometry: Complex Numbers (5 lessons, $6.93)

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/collegealgebra. The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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Okay, adding complex numbers, subtracting complex numbers, not a big deal. Multiplying complex numbers just requires a little bit of foiling. What about dividing complex numbers? Well, in fact, there's a little bit of trickery in here, too, which, once you see the idea, it's not a big deal, but it does require something. Let's take a look at an example.
Suppose I wanted to consider . Well, you know, to inspire this, let me actually remind you of what this really is. You have to remember what i is here, so I have . Aha! Square roots on the bottom and you may recall that there are some prejudicial people in mathematics who just don't lie that. So they want to rationalize the denominator. And that would actually require multiplying by a very clever choice of 1, in this case, something over itself and, in this particular case, the trick remember is to multiply top and bottom by the conjugate of the denominator. And the conjugate is the exact same thing as it was before, the same thing as this number, except this sign is changed. So if you multiply top and bottom by the conjugate, what happens is this becomes a good, old-fashioned real number, as we will see, because the inner terms will cancel out. So the trick in dividing complex numbers is going to be to multiply top and bottom by the conjugate of the denominator.
So how would that look? I would have 3 - 5i, that's the conjugate of the bottom. It doesn't change the value, this is just 1. Now, I admit - did you hear that cough, by the way, that was our camera person, Sam. It's a girl. Isn't that sort of funny, because you think Sam, anyway... So this cancels out and it's just 1. So it doesn't change the value. So now, what happens when I multiply this through? I have to multiply the tops and I have to multiply the bottoms. So this is going to be a whole bunch of foiling, like we just talked about, in terms of multiplying complex numbers. So let's take a look at how that would work.
Well, I'd have the 4 times the 3, which produces a 12, and them my inside term produces the complex number 3i. You may start doing this all in one step, but I'm not going to do that right now. And then the outsides produce the complex number of -20i. And then the last times the last, now I'm going to do this in one swell swoop. It looks like -5i2, but remember i2 = -1. So I actually see -5 -1, which is a plus 5. So that takes care of the top.
What about the bottom? The bottom I see 3 3, which is 9. The inside term is 15i. The outside term, notice, is -15i. So 15i - 15i, they cancel. That was the whole point of this, by the way. And the last times the last is -5i2, but remember i2 = -1, so I see -5 -1, which is a plus 5. Notice the denominator now is a real number. That was the whole point of this exercise is to divide this into it and have no i's in here. So I multiply top and bottom by the conjugate.
Let's now reduce this a little bit, if we can simplify it and put it together. The real parts that produce a 17, the imaginary parts, -20i + 3i = -17i, divided by - and then, let's see, let's make sure I did it correctly. So I've got 17 and then if I take 20 and I subtract 3, let's see what happens here. Now, there's a problem, let's see if we can figure it out here. No, there's no problem here, but look what I wrote here. You might have seen this even earlier on. Oh, boy! You know, making mistakes is fine, but you've got to always be careful. I took this times this and I talked through it and I said that - this is -25 i2. I think I said -5. I forgot to multiply this by this. Sorry about that. This is wrong, this is completely wrong. You know what I'm going to do, in fact? I'm so embarrassed by this that I don't want you to even look at that anymore. Let's just cover it up. I don't even want to patch that up. I just want you to not even think about it, because that was wrong. It shouldn't have been 5, it should be 25, because it's this times this, which is a -25i2, which is plus 25. But it's good for you to see that everyone makes mistakes, especially me.
The top is fine, and if I take 9 and I add 25, I get something like 34? Oh, I can factor a 17 out of the top and I'm left with . And look, 34, I can cancel the 17 and the 34 and that leaves me with , which I could write this way. Some people like to write these things this way, because then you can see the real part is and the imaginary part is . So you could break up that fraction and write it that way.
Okay, let's try another one, so you can get the hang of this. In fact, let me let you try this one, . I want you to do that division, see what you can get. Remember the trick is to multiply by the conjugate. I wonder what the conjugate would look like here? Okay, you try.
Okay, well, what can you do here? If you multiply top and bottom by the conjugate, you could think of this as actually 0 + i. So you'd want to multiply top and bottom by 0 - i and that would be the conjugate. In fact, in this problem, it turns out to be so easy. You can just multiply top and bottom by i, if you wanted to, but just to be consistent, even though that would be perfectly fine and, if I were in private, I would do that, I'll multiply top and bottom by the conjugate.
So the conjugate of i is actually -i. And if I do that, what do I see? Well, I have to distribute now. Everything has to get hit with that thing. I don't need to foil now, because it's such a simple thing on top, so I'd see -6i, and then, let's see, a minus and a plus is a -i2, so a -i2 is a minus -1, which is a plus 1. And on the bottom I'm left with an i i = i2, so that's -i2, but -i2 is minus -1, which is 1. So, in fact, this just turns out to be 1 - 6i. So that one was not too bad to compute. Let's try one last one together.
Let's find the reciprocal of 2 + i. So that means I'm looking for . How would you do that? Well, I'll give you a chance to do it right now. So see if you can figure out what the reciprocal of 2 + i is.
Well, the key is to just multiply top and bottom by the conjugate of the denominator. It won't change the value, but it will allow us to figure out what that number is. So that would be 2 - i, top and bottom. On the top, that's pretty easy. It's just 2 - i. On the bottom 2 2 = 4. Then I have an inside term of 2i and an outside term of -2i and they cancel. That's always supposed to happen. If you're multiplying by the correct conjugate and do the multiplication correctly, you should, in fact, always get those middle terms, outside and such terms, to drop out. The last term is going to be a -i2, which is going to be plus 1. So we see . That's fine, but if you really want to be careful with the real part, you can break that up into two little fractions. It would look like this; you would see , because if you combine them, the 5's are the same, I just combined the tops. And so I see 2 - i.
So, in fact, the reciprocal of 2 + i is this. That's sort of weird. What that means is if I take this number and multiply it by 2 + i, I should get 1, because this is the flip of this. And you do, and you can work it out and try it.
Anyway, that's all there is to dividing and conquering numbers. Just remember that, when they're complex, multiply top and bottom by the conjugate. You know what? It's time for me to go to lunch. See you.
Prerequisites
Complex Numbers
Dividing Complex Numbers Page [2 of 2]

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