Complex Numbers and the Mandelbrot Set

Date: 12/15/96 at 23:41:39
From: Kevin Ong
Subject: Fractals

I had to choose a subject on fractals for aproject. I chose the 
Mandelbrot set and how it works.  I don't understand the equation for 
the Mandelbrot set fractal.  I know that the equation is z = z^2 + C.  
I don't understand where you get the numbers to fill in the variables 
and how you would graph this.

Date: 12/16/96 at 15:39:59
From: Doctor Pete
Subject: Re: Fractals

Hi,

To understand the Mandelbrot set, one first needs to be familiar with 
complex numbers.  If you know the basic arithmetic of complex numbers, 
then you can skip the next few paragraphs.  If not, read it carefully.

Complex numbers were introduced to answer the question of "What is the 
solution to x^2 + 1 = 0 (x squared plus 1 equals 0)?"  Now, you'd 
think that such an equation would have no solution, and this is true 
if x has to be a *real* number.  But let's call i = Sqrt[-1] (the 
square root of -1).  Then x = +i and x = -i are solutions to the above 
equation.  And you're thinking, "But you're not allowed to take the 
square root of a negative number!"  Well, this is not completely true.  
The "rule" should say, "The square root of a negative number is not a 
real number."  We don't mind the fact that this new number i is an 
apparent contradiction because in defining it, we give it the 
following properties:

i follows the "regular" operations of addition, subtraction, 
multiplication, and division as for the reals, and i^2 = -1.

So i is what we call the *imaginary unit* of the *complex* numbers.  
In the reals, there are two units:  1 and -1.  In the complex numbers, 
which are numbers of the form a + bi (a plus b times i) for real 
numbers a and b, there are four units:  1, -1, i and -i.  So the 
complex numbers are an extension of the reals, since when b = 0, 
we get the real numbers.  Addition of two complex numbers gives a 
complex number:
  
(a+bi)+(c+di) = (a+c)+(b+d)i
 
Multiplication of two complex numbers also gives a complex number:
  
(a+bi)(c+di) = ac + adi + bci + bdi^2 
             = ac + (ad-bc)i - bd = (ac-bd) + (ad-bc)i
               (remember, i^2 = -1)  

Just as with the reals, there is an additive identity (0 + 0i = 0), 
and a multiplicative identity (1 + 0i = 1).  Every nonzero element 
a+bi has an additive inverse, namely -a-bi, and a multiplicative 
inverse, namely (a-bi)/(a^2+b^2).  So when we multiply, add, divide, 
or subtract two complex numbers, the result is also a complex number 
(except when dividing by 0, of course, which is still undefined).  
Sometimes, though, we simply write complex numbers as a single 
variable, usually z or w.  That makes it easier to work with in some 
cases.

Hopefully, you're familiar with the real number line.  It's a 
graphical representation of the real numbers, and is usually depicted 
like this:

    <----|---|---|---|---|---|---|---|---|---|---|---->
    ... -5  -4  -3  -2  -1   0   1   2   3   4   5  ...

Every real number is a point on this line.  Now, the complex numbers 
have a graphical representation as well, but it's a lot more 
interesting.  It's the *complex plane*.  Every complex number is a 
point in a Cartesian coordinte system, where the x-axis is the real 
axis, just like the real number line, and the y-axis is the 
*imaginary* axis, where complex numbers of the form 0+bI are plotted.  
It looks like this:

                             ^ Im(z)
                             |
                          3i +
                             |          3+2i
                          2i +           .
                             |
                           i +
                             |
           <-+---+---+---+---+---+---+---+---+->
            -4  -3  -2  -1   |0  1   2   3   4
                          -i + 
                             |
                         -2i +       . 2-2i
                             |
                         -3i +   .
                             |  1-3i
                             +

Here I have plotted the points 3+2i, 2-2i, and 1-3i.  Fairly 
straightfoward stuff, especially if you've had coordinate geometry.

Now, what does this have to do with the Mandelbrot set?  Well, you 
mentioned the equation z = z^2 + c, but a more accurate way of writing 
this is:

     f(z) :  z |--> z^2 + c

That is, we have a function f, where z maps to z^2 + c, or for a value 
of z, the equation tells us to square it and add c to it.  This in 
itself is not all that interesting, but what happens if we keep 
applying f.  For example,

     f(0) = c
     f(f(0)) = c^2 + c
     f(f(f(0))) = (c^2 + c)^2 + c
     f(f(f(f(0)))) = ((c^2 + c)^2 + c)^2 + c
     . . . . . . etc.

The French mathematician Gaston Julia asked the question, "For a 
given value of c, for what values of z does f(f(f(...f(z))) stay 
bounded?"  In other words, we pick a value of c, say c = 0, and we 
pick a value of z, say z = 1/2.  Then we calculate:

     f(1/2) = (1/2)^2 = 1/4
     f(1/4) = (1/4)^2 = 1/16
     f(1/16) = (1/16)^2 = 1/256

Clearly, if we keep going, z will get smaller and smaller, approaching 
0.  Now, what if we picked z = 2?  Then f(2) = 4, f(4) = 16, 
f(16) = 256, and it is just as clear that z will increase without 
bound.  If we pick z = 1, then f(1) = 1, and no matter how many times 
we apply f, z stays fixed at 1.  So on the positive real line, all the 
points between 0 and 1 inclusive stay bounded under this mapping, and 
all other points do not because they increase without limit.  If we 
define Or+(f,z) to be the *forward orbit* of z under f (the infinite 
set {z, f(z), f(f(z)), f(f(f(z))), ...}), then Or+(f,z) is *bounded* 
if a finite region contains every point in Or+(f,z).  We call each 
element in Or+(f,z) an *iterate* of z under f.  In particular, if we 
repeat f on z n times, f(f(...f(z))) is the n(th) iterate of z under 
f.  So Or+(f,1/2) is bounded, since every element/iterate in it is 
contained in the interval [0,1/2], and Or+(f,2) is not bounded, since 
you cannot pick a *finite* interval that contains all iterates.

Well, this is pretty straightfoward, but what happens when we pick 
different values of c?  Well, this is what Julia asked, but he gave 
things a twist - what if z and c were *complex* numbers, rather than 
merely *real* numbers?  It turns out that this leads to very 
interesting results, so interesting that they are still being 
investigated a hundred years after Julia asked the question.

We define a *Julia set* to be the set of all complex numbers z such 
that Or+(f,z) is bounded, where f(z) : z |--> z^2 + c, and c is some 
given complex number.

I leave it as an excercise for you to show that if c = 0, the 
corresponding Julia set is a disk of radius 1 when plotted in the 
complex plane.

Now, Benoit Mandelbrot, sometime in the late 70's/early 80's, asked 
the question, "For what values of c is Or+(f,0) bounded?"  That is, we 
pick various values of c, and calculate f(f(f(...f(0)))), and find out 
if this stays bounded or diverges to infinity.  The rather surprising 
result is that when one plots the various values of c (*not z*, but c) 
in the complex plane, one obtains a fractal.

We define the *Mandelbrot set* to be the set of all complex numbers c 
such that Or+(f,0) is bounded, where f(z) : z |--> z^2 + c.

There are unusual properties of the M-set, some of which are very 
difficult to prove:

1) The M-set is "connected," that is, for any two points in the 
M-set, there is a continuous line connecting those two points.

2) The M-set is the set of all c for which the corresponding Julia 
set is connected.  That is, if a point c is in the M-set, then the 
Julia set for that value of c is connected.

3) The M-set is contained in a disk of radius 2, and a point is *not* 
in the M-set if there exists some iterate in Or+(f,0) which is outside 
this disk.  (This gives an easy way to check that a point is not in 
the set.)

Well, that concludes my overview of the Mandelbrot set.  I've written 
a lot here, and depending on your level of mathematics, there is a 
great deal more that can be said.  I will leave it to you to research 
the topic further.  Check out your local library; you'll find a lot of 
accessible books on the topic, as fractals have become quite popular 
in the "armchair science" community.  Mathematicians have been 
studying them for decades.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/