Defining Complex NumbersDate: 12/01/97 at 16:24:48 From: Anonymous Subject: Complex numbers Hi. Complex numbers seem very confusing. Is there a simple defination or a way to explain them so a student like me can understand them? Date: 12/01/97 at 19:12:26 From: Doctor Pete Subject: Re: Complex numbers Hi, Complex numbers are usually taught like this: "Let i be the value such that i^2 = -1...." But this is hard for many students to grasp at first. On the other hand, I like to introduce complex numbers without using square roots of -1. Instead, think of ordered pairs of real numbers, like (3,5), (-1,0), or (5/2,3.14). In general, let (a,b), (c,d) be such ordered pairs. We now *define* a structure on these pairs, specifically, the operations of addition and multiplication. For addition, (a,b) + (c,d) = (a+c,b+d) Notice that (a+c,b+d) is also an ordered pair of real numbers. Multiplication, however, is defined as (a,b) * (c,d) = (ac-bd,ad+bc) So we see that the product of two ordered pairs is also an ordered pair of real numbers. Notice that (a,b) + (0,0) = (a+0,b+0) for any (a,b), so (0,0) behaves much like 0 in the real numbers. Also, (a,b) + (-a,-b) = (a-a,b-b) = (0,0), so (-a,-b) is like the "negative" of (a,b). Now, you might think that (a,b) * (1,1) = (a,b) but this is not true! Using the definition, (a,b) * (1,1) = (a-b,a+b) so (1,1) is NOT like 1 in the real numbers. Instead, we see that (a,b) * (1,0) = (a*1 - b*0, a*0 + b*1) = (a,b) so in fact, (1,0) is our "1" in this new multiplication. So you can see that pairs of the form (a,0) behave just like the real numbers - it's just that I've attached this extra "0". But then according to our definition, (0,1) * (0,1) = (0*0 - 1*1, 0*1 + 1*0) = (-1,0) So (0,1)^2 = -1. In other words, this "extra attachment" does *NOT* have the same properties as the real numbers, or else you would have expected (0,1)^2 = 1. There are a few minor details left: (a,b) * (a,-b) = (a*a - b*(-b), a*(-b) + b*a) = (a^2 + b^2, 0), so (a,b) * (a/M,-b/M) = (1,0), where M = (a^2 + b^2). In other words, (a/M,-b/M) is like our "reciprocal," as 1/5 is the reciprocal of 5. It's the number that when multiplied by a given number gives 1. So we can now add, subtract, multiply, and divide, because (a,b) - (c,d) = (a,b) + (-c,-d) and (a,b) / (c,d) = (a,b) * (c/M,-d/M) where M = (c^2 + d^2). Pretty straightforward, right? Notice that I don't say anything about square roots of negative numbers. Now, the definition of the complex numbers, is this: Definition: The *complex numbers*, or C, is the set of ordered pairs (a,b), where a and b are real numbers, such that addition and multiplication are defined as above. That's basically all there is to it. Think of complex numbers as ordered pairs, so (3,5) is a complex number. For example, (3,5) + (17,-2) = (20,3), (3,5) * (-2,7) = (-6-35,21-10) = (-41,11). The final twist: Instead of writing things as, say, (3,5), notice that we can write them (3,0) + (0,5). Taking this one step further, write 3(1,0) + 5(0,1). But (1,0) is just like our real number 1, so write 3 + 5(0,1). Unfortunately, we don't have a "name" for (0,1). It has no "partner" in the real numbers, so what mathematicians have done is to say (0,1) = i. But as we have said, (0,1)^2 = -1, which is why i^2 = -1. Now, i is not a variable; it is clearly a *number*, but written this way, we can now say (3,5) = 3 + 5(0,1) = 3 + 5i And in fact, this is nicely generalizable for any ordered pair: (a,b) = a + b(0,1) = a + bi Notice that we treat the parts that don't have i's attached separately from those that do; the plus sign between the a and the bi is a bit confusing, but there is no confusion when we clearly separate them by the comma: (a,b). (a,b) and (b,a) are not equal or interchangeable. So how do we do addition, multiplication, with this new notation? (3+5i) + (17-2i) = 20+3i. (3+5i) * (-2+7i) = -6+35(i^2) + (21i - 10i) = -6-35 + i(21-10) = -41 + 11i, since i^2 = (0,1)^2 = -1. It's more confusing at first, true. The ordered pairs make it easy to separate things. But with practice and time, you'll get the hang of it; it's just a different kind of notation. One last thing: What is Sqrt[-4]? You've probably been asked this. Here's how to think about it with ordered pairs: you want to find an ordered pair (a,b) such that (a,b)^2 = (-4,0). But (a,b)^2 = (a*a - b*b, a*b + a*b) = (a^2 - b^2, 2ab). If we require this to be (-4,0), then a^2 - b^2 = -4, 2ab = 0. The second equation means either a, or b, or both, is 0. But the first equation tells us that one of them must be nonzero. Since a and b are real, and the square of any real number is nonnegative, clearly we must have a = 0. Otherwise, if b = 0, then a^2 = -4, which has no solution for real a. Therefore, we conclude a = 0, -b^2 = -4, or b^2 = 4, or b = 2. (-2 works, too!) So we find that (0,2)^2 = (-4,0), and Sqrt[-4] = (0,2) = 2(0,1) = 2i. You can see that Sqrt[-n] = Sqrt[n](0,1) = Sqrt[n]*i for nonnegative real n. Wow, that was a lot of stuff.... I hope you got the hang of it; if not, take some time to absorb it, see if it works for you. The important things to remember are thinking of complex numbers as a *pair* of reals, with addition and multiplication *defined* as I defined them at the beginning. The rest naturally follows. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/