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Defining Complex Numbers


Date: 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/   
    
Associated Topics:
High School Imaginary/Complex Numbers

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