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Imaginary (Complex) Numbers

Date: 02/26/98 at 10:00:18
From: Chris
Subject: Imaginary (complex) numbers

What are imaginary numbers?  I have to do a report on them and can't 
find anything understandable anywhere else.  Thank you for your time.

Date: 02/26/98 at 15:25:10
From: Doctor Rob
Subject: Re: Imaginary (complex) numbersz

You probably are aware that for any real number a, a^2 >= 0. If one is 
faced with an equation like x^2 = -9, one can immediately conclude 
that there is no real solution.  Imaginary numbers were invented to 
provide solutions of equations like this.

The simplest imaginary number is called i. It satisfies the property 
that i^2 = -1. Using it, the solutions of x^2 = -1 can be found:  
x = i is one, and x = -i is another, if we define -i properly. Since i 
cannot be a real number, it must be something else. It is called an 
"imaginary" number. Of course, it is no more imaginary than 7, 0, -4, 
3/5, sqrt(2), pi, or any other number. They are all figments of our 

In order to be a "number," i must satisfy the axioms of arithmetic.  
It must have an additive inverse, or negative, or opposite, -i, so 
that i + (-i) = 0. It must have a multiplicative inverse, or 
reciprocal, which it does: in fact -i = 1/i. (You can prove this 
by showing that i*(-i) = 1.) Thus -i is both the negative and the 
reciprocal of i. Multiplication by a real number must be defined, 
so a*i must make sense, for any real number a. In particular, 
(-1)*i = -i. All the other axioms of arithmetic must also be true 
for the set containing the real numbers and i and whatever else you 
can get by doing arithmetic with them.

Now you can solve x^2 = -9 = 9*(-1) = 3^2*i^2 = (3*i)^2. The two 
solutions are x = 3*i and x = -3*i, imaginary numbers! The same trick 
works for x^2 = -a, where a > 0 is a real number.  x^2 = -a = a*i^2, 
so x = sqrt(a)*i and x = -sqrt(a)*i are the two solutions, imaginary 

The system of numbers you end up with by looking at the set containing 
the real numbers and i and whatever else you can get by doing 
arithmetic with them has a name, of course. It is called the complex 
numbers. It turns out that every complex number can be written in the 
form a + b*i, for some real numbers a and b. This means that every 
complex number is the sum of a real number and an imaginary number.  
The rules for doing arithmetic with complex numbers are all the 
ordinary ones, because we insisted that that be so when we built up 
the set. The only difficulty in seeing that the form a + b*i includes 
all complex numbers is in multiplication and division. For 
multiplication, do the ordinary multiplication of two complex numbers
of this form using FOIL, then use the fact that i^2 = -1, and combine 
terms containing i and ones not containing i.  For example,

   (3 - 2*i)*(-1 + i) = -3 + 3*i + 2*i - 2*i^2,
                      = -3 + 3*i + 2*i + 2,
                      = -1 + 5*i.

The result is back in the same form, a real number -1 plus an 
imaginary number 5*i.

For division, we have to know one trick: 
(a + b*i)*(a - b*i) = a^2 + b^2.
Multiplying these two complex numbers gives a real number answer!  
If you find this hard to believe, multiply it out using FOIL and 
i^2 = -1, and see what you get.

Now we can do division. To divide by a + b*i, instead multiply by its
reciprocal, which is

   (a - b*i)/(a^2+b^2) = a/(a^2+b^2) + [-b/(a^2+b^2)]*i,

and use the rules given above to reduce it back to the form of a real
number plus an imaginary number. Of course, we cannot do this if
a^2 + b^2 = 0, but that only happens if a = b = 0, so the divisor is
0 + 0*i = 0. Division by zero is forbidden, but we are used to that!

As an example, since (-1 + i)*(-1 - i) = (-1)^2 + 1^2 = 2, and so the
reciprocal of -1 + i is (-1 - i)/2 = (-1/2) + (-1/2)*i, we get

   (3 - 2*i)/(-1 + i) = (3 - 2*i)*(-1 - i)/2,
                      = (-3 - 3*i + 2*i + 2*i^2)/2,
                      = (-5 - i)/2,
                      = -5/2 - (1/2)*i.

Now you can do any arithmetic with complex numbers, and in particular, 
with imaginary numbers.  You can also take square roots of any real 
number, whether positive, zero, or negative.

-Doctor Rob,  The Math Forum   
Associated Topics:
High School Number Theory

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