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### Complex Numbers and the Mandelbrot Set

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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
(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

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Fractals
High School Imaginary/Complex Numbers

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