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### Defining Complex Numbers

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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?
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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

(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

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/
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Associated Topics:
High School Imaginary/Complex Numbers

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