module on complex numbers

7/27/2019 Module on Complex Numbers

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Lesson 1: Basic Understanding and

Operations of Complex Numbers (slides

?-?)

Lesson 2: Geometric understanding of

addition and subtraction (slides ?-?)

Lesson 3: Applications (slides ?-?)

Module on Complex Numbers


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Click on the picture to watch the

video

Video Annenberg
http://localhost/var/www/apps/conversion/tmp/scratch_7/COMPLEXVIDEO.htmlhttp://localhost/var/www/apps/conversion/tmp/scratch_7/COMPLEXVIDEO.htmlhttp://localhost/var/www/apps/conversion/tmp/scratch_7/COMPLEXVIDEO.html


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Introduction:Why do we need new numbers?

The hardest thing about working with complex numbers

is understanding why you might want to. Before

introducing complex numbers, let's backup and look at

simpler examples of the need to deal with new numbers. If you are like most people, initially number meant whole

number, 0,1,2,3,... Whole numbers make sense. They

provide a way to answer questions of the form "How

many ... ?" You also learned about the operations ofaddition and subtraction, and you found that while

subtraction is a perfectly good operation, some

subtraction problems, like 3 - 5, don't have answers if

we only work with whole numbers.

Lesson 1: Basic Understanding and Operations of Complex

Numbers

3


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Then you find that if you are willing to work

with integers, ...,-2, -1, 0, 1, 2, ..., then all

subtraction problems do have answers!

Furthermore, by considering examples suchas temperature scales, you see that negative

numbers often make sense.

Now that we have fixed subtraction we willdeal with division. Some, in fact most, division

problems do not have answers that are

integers. For example, 3 2 is not an integer.

We need new numbers! Now we have rationalnumbers (fractions).

There is more to this story. There are

problems with square roots and other

operations, but we will not get into that here. 4


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The "problem" that leads to complex numbers

concerns solutions of equations.

What if we want to find a number that when multiplied

it by itself equals -1?" Or can you solve the equation x2 + 4 = 0 for x?

****************************************************

The need to extend the real numbers was prompted

by the desire to solve problems like the following,

which appears in the 1545 book Ars Magna by

Cardana:

Divide 10 into two parts whose product is 40.

To solve this problem, Cardano needed to solve the

equation

x(10-x)=40, which is equivalent to x2-10x+40=0.

The solution of the equation requires to be a

number. But -15 does not have a square root that is a

15

5


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In the real number system, the square root of a

negative number does not exist. That's because

there are no real numbers whose squares are

negative. However, you can take the square root of a

negative number if you are willing to use a non-

real number to do it. This new number was

invented (discovered?) more than 400 years ago.It was called "i", standing for "imaginary",

because iwasn't "real".

Imaginary number :The imaginary numberi is defined to be i 1

Then i 22

1 1

But beware of the following: i 22 2

1 1 1 1 6


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But ialready squares to1. So it cannot also squareto 1. This points out an important detail: when dealingwith imaginary numbers, you gain something (theability to deal with negatives inside square roots), but

you also lose something (some of the convenientrules you have when dealing with square roots).Lacking these rules, we make the following definition:

Ifxis a positive real number, then

Examples:

You can check our answers on a calculator which

permits complex number arithmetic. The TI-84 is onesuch calculator, and we will use it throughout thismodule. After turning it on, press the MODE key.Move down to REAL and over to a + bi. With thecursor blinking over a + bi, press ENTER to savethis complex mode of arithmetic. Now go back to thehome screen by pressing 2ND QUIT. You are ready

x i x

9 3 18 3 2i i

7


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Powers of i:

We have seen that . It then follows thati 2 1

3 2

4 2 2

5 4

6 2 4

7 3 4

8 4 4

1

( 1)( 1) 1

1

1 1 1

1

1 1 1

i i i i i

i i i

i i i i i

i i i

i i i i i

i i i

8


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Notice that the powers oficycle around four numbers:

i, -1, -i, 1

If the exponent is a multiple of 4, the power equals 1.

For example,

To calculate any high power ofi, you can convert it to alower power by taking the closest multiple of 4 which isno larger than the exponent and subtracting thismultiple from the exponent. For example,

You can confirm these answers by calculator: The i

key is found in the middle of the bottom row. Tocompute ito the power 27, press 2ND i^27 ENTER.Your answer should bei, but perhaps you instead gotsomething like -3E-13 - i. What is going on here? Thereal part of this result is -3 times 10 to the power -13,which is within machine roundoff error of zero. So, you

i 24 1

i i i i i i27 24 3 24 3 3

9


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Pause and practice-1

Simplify

First compute these by the rules above, then

check answers on your calculator.

55 62 73( ) ( ) ( )a i b i c i

Answer: (a)i (b) -1 (c) i

10


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Complex numbers:

What if we want to combine a real number with animaginary number? We could say that 3 + 4i is a

number but it has more parts to it than a normal

number. We call it a complex number.

Every complex numbercan be written in the form a +bi, where a and b are real numbers, called the real

partand the imaginary partof the complex number,

respectively.

For example, 2 + 3iis a complex number, with realpart 2 and imaginary part 3.

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It's as though our imaginary number 'i' isn't on the

number line, but we must be able to put it

somewhere. And what about 2i and 3i and -7i?

We must be able to put them all somewhere. Why

don't we make our imaginary number line

perpendicular to the real number line through the

origin. then not only will we have a place for

imaginary numbers like 5i and -3i but also forcomplex numbers like 2+4i and -2 - 5i. Its like we

now have a visual way for looking at complex

numbers.

A complex number can be visually

represented as an ordered pair of numbers

specifying a point in thexy-plane. This point

is the tip of a vector emanating from theori in and this vector also re resents the


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Below is a graph of the vector representing the

number

2 + 3i. Note that the tip of the vector has coordinates

(2, 3).

In general, any complex numbera + bican be plotted in

thexy-plane (also called the complex plane) as the

point having coordinates (a, b). This point is the tip of

a vector emanating from the origin. Important special

cases: a pure real number is plotted on thex-axis, 13


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Arithmetic of complex numbers:

Given two complex numbers, we now define howto add, subtract, multiply, and divide them. We

want to do this in a natural way so that the usual

rules for arithmetic of real numbers continue to be

valid for complex numbers. In particular, we wantaddition and multiplication to be commutative,

associative, and distributive. Since the complex

numbera + bilooks a bit like the linear

polynomial a + bx, lets use our knowledge ofpolynomials as motivation. So, to add, subtract,

or multiply complex numbers, we simply combine

like terms as is seen in the following examples:

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Operations:

Addi t ion example:

Subtract ion examp le:

APPLET-1

15

( ) ( )

( ) ( )

2 4 3

2 3 4

5 3

i i

i i

i

( ) ( )

( ) ( )

2 4 32 4 3

2 3 4

1 5

i ii i

i i

i
http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/addition_and_subtraction_of_complex_numbers_.htmlhttp://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/addition_and_subtraction_of_complex_numbers_.html


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Operations:

Mult ip l icat ion example:

Note that we replaced by -1 in the above

calculation.

( )( )

( )

2 4 3

6 2 12 4

6 10 4 1

6 4 10

10 10

2

i i

i i i

i

i

i

APPLET-2

16
http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/multiplication_and_division.htmlhttp://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/multiplication_and_division.html


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Add, subtract, and multiply the complex numbersand .

First do these problems by hand, then check your

results by calculator.

5 i 2 3 i

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What about d iv is ion?

If the denominator is pure imaginary, we canmultiple by to eliminate the iin the

denominator, as is seen in this example:

This was simple enough, but what if you have

something more complicated, such as

In order to eliminate the iin the denominator, we

make use of the conjugate.

i

i

1 11i i

i

i

ii

5 23 4

i

i

18


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The con jugate of a complex number a

+ bi:

The conjugate of a complex numbera + biis thesame number, but with the opposite sign in the

middle: a bi. For example, the conjugate of 3 +

4i is 3 4i.

The multiplication by conjugates produces a sumof squares. You should pause to verify this fact:

This is similar to the more familiar difference of

squares:

( )( )a bi a bi a b 2 2

( )( )a b a b a b 2 2

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Division examp le:

Note that in the first step we multiplied both numeratorand denominator by the conjugate of the denominator.

In the last step, note how the fraction was split intotwo pieces. This is because, technically speaking, acomplex number is expressed as a sum of two parts:a + bi.

To check this example on your calculator, enter

(5+2i)(3+4i) to get .92 - .56i. If you prefer fractions

5 2

3 4

5 2

3 4

3 4

3 4

15 20 6 8

9 12 12 16

23 14

2523

25

14

25

2

2

i

i

i

i

i

i

i i i

i i i

i

i

20


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Pause and Practice -3

Divide: (3 + 4i) (5 2i) by hand, and thencheck your answer by calculator.

21


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Modulus (o r, Abso lute Value): We have seen how complex numbers can be

represented by points in the complex plane.

Unlike the real numbers, there is not a natural

way to order them. For example, 3 < 4, but how

would you compare 3 + 2iand 4 + i? We do this

by computing the distance each point is from the

origin. Using the distance formula,

The modulus (or, absolute value ) of a complexnumbera + biis defined to be its distance to the

origin, and is denoted by

Note that this definition agrees with the definition

3 2 13 4 1 172 2 2 2 ,

a bi a b 2 2

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In our example, we would say that the modulus

of 3 + 2i is less than the modulus of 4 + i.

You can check your answers on the calculator

using abs(a+ib), as seen in the following

example:

Select MATH NUM abs(3+2i) ENTER, giving a

decimal approximation to the square root of 13.

Since this number is irrational, you cannotconvert it to a fraction using FRAC.

23


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For each of the following complex numbers, plotthe complex number as a point in the complex

plane, connect the point to the origin by a

segment, and compute the modulus (first by

hand, then by calculator) to obtain the length ofthis segment.

(a) 3 + 4i (b) -5 + 12i (c) 8 6i

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Lesson 1 - Quiz

Try to answer each question twice,without a calculator and then with a

calculator.

1.

2.

25


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Lesson 1 - Quiz

3. Draw the vector representing 2-i .

4.

5.

26


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Lesson 1 - Quiz

6.

7.

Answers: 1(b), 2(a), 4(c), 5(a), 6(b), 7(d)27


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Introduction:

In this discussion, it will be convenient to denotea complex number by a single variable name.

??

Lesson 2: Geometric understanding of addition and

subtraction

28


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Let w= a + bi and z= c + di.

If O denotes the origin, then the following four

points form vertices of a parallelogram: O, w, z,

and w + z.

For example, ifw = 2 + iand z = 1 + 2i, then

w + z= 3 + 3i. It is easy to see that (0, 0), (2, 1),

(1, 2), and (3, 3) form vertices of a parallelogram.The points (0, 0) and (3, 3) are endpoints of one

diagonal, while (2, 1) and (1, 2) are endpoints of

the other diagonal.

In general, O and w + z areendpoints of one diagonal,

while wand zare endpoints of

the other diagonal.

Please click the button 29

w

z

w+z

Applet

for

addition
http://www.walter-fendt.de/m14e/complnum.htmhttp://www.walter-fendt.de/m14e/complnum.htmhttp://www.walter-fendt.de/m14e/complnum.htm


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Sketch the parallelogram formed by the complexnumbers

w= 1 + 2i, z= -2 + I , origin and w+z.

Is this a rectangle? a square?

Also confirm it with the applet

30

Appletfor

addition


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What about subtraction?

Once again, a parallelogram is formed: by thefour points: O, w, z, and w z. However, in this

case O and ware endpoints of one diagonal,

while zand w - zare endpoints of the other

diagonal.

Try with the applet

31

Applet for

subtractio

n


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O and ware endpoints of one diagonal, while zand w - zare endpoints of the other diagonal.

Confirm this fact in the case where w= 1 + 2iand

z= -2 + i.

And check with the appletApplet forsubtraction

C l b i l
http://www.walter-fendt.de/m14e/complnum.htmhttp://www.walter-fendt.de/m14e/complnum.htmhttp://www.walter-fendt.de/m14e/complnum.htmhttp://www.walter-fendt.de/m14e/complnum.htm


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Complex numbers in polar

form

In order to gain a geometric understanding ofmultiplication, it will be convenient to represent points

in the complex plane using polar coordinates.

Suppose that z = a + bi. Rather than represent this

point using rectangularxy-coordinates, we can usepolar coordinates (r, ). The variable rrepresents the

distance from point zto the origin, and is the angle

(measured counterclockwise) the vectorzmakes with

the positivex-axis.

33


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Using the definitions of the trigonometricfunctions sine and cosine, we have

Therefore, zcan be written

The expression arises so frequently that it iscustomary to abbreviate it as cis . Then, z=rcis is called thepolar form of a complex

number. Since the distance to the origin is represented by

the modulus ofz, we have . The angle is called the argumentofz. Note that the

argument can have many possible values, but

cos ,

sin .

a r

b r

z a bi

r ir

r i

cos sin

cos sin

r z

34


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There is an interesting connection between the

polar form of a complex number and the real

function , which is the inverse function of

the natural logarithm. In the 18th century,Leonhard Euler discovered that

, provided that is measured in radians. Thus,

and so the polar form of a complex

number can be written as

The polar form is available on the TI-84

calculator. Simply press MODE, then move the

cursor down to REAL and over to re^ , thenpress ENTER. The value of is expressed in

radians or degrees, depending on the MODE

setting.35


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Example: Find the polar form of z = 1 +

i.

The vector represented by zis the segment fromthe origin to the point (1, 1). By drawing a sketch,one easily sees that the argument is 45 degrees (multiples of 360). In cases which are not clear,one can use the formula . In our example,bothxand yhave value 1.

To obtain the modulus, .Thus, the polar form is

You can check this answer on your calculator bytyping:

1 + iENTER. Of course, your MODE settingshould already be set to polar: re^

tan y

x

2 21 1 2r z

2cis 45z

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Example: Find the standard form of z= 2 cis 150.

We make use of the formulas

This gives

Thus,

We now check this answer on the calculator. Sincethere is no cis key, we make use of the key 2ND .

Also, recall that Eulers formula is valid

only in radians. So, we first must convert 150

degrees to radians and enter our expression asENTER. This gives the desired result, but

with an approximation to . Note that 2e^(150*i)

ENTER gives a very different (and wrong!) answer.

a r

b r

cos ,

sin .

2 cos 150 3 and 2 sin 150 1.a b

3z i

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Try these by hand, and then check with yourcalculator.

(a) Find the polar form of z = 43i.

(b) Find the standard form of z= 2 cis 225.

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Geometric understanding of

multiplication

We now show that the product of two complexnumbers has modulusequal to the product of the

individual moduli and argument equal to the sum of

the individual arguments.

Let w= |w| cisA and z= |z| cis B. Then,

wz= (|w| cisA)( |z| cis B)

= |w||z|(cosA + isinA)(cos B + isin B)

= |w||z|(cosA cos B sinA sin B + isinA cos B + icosA sin B)

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Using the trigonometric angle sum formulas, thislast expression can be written as

= |w||z|(cos(A + B) + isin(A + B)).

Thus, the polar form ofwzis

wz = |w||z|(cos(A + B) + isin(A + B)),

which allows us to identify |w||z|as the modulus ofwz, and

A + B as the argument ofwz.

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Let w = 2i, and let z = 1 + i.(a) Find the product wz.

(b) Find the modulus of each ofw, z, and wz, andthen try to decide how these three moduli are

related to each other. Relate your answers to thepolar forms of all three numbers (Check theirpolar forms on your calculator.)

(c) Plot the three points w, z, and wzin the complex

plane, and connect each point to the origin with aline segment. Try to decide how the anglesbetween the positive real axis and these threelines are related to one another. Relate youranswers to the polar forms of all three numbers

(Check their polar forms on your calculator.) 41


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De Moivres Theorem:

We can use the polar form to square a complex number:whereA is the

argument ofz.

whereA is the argument ofz. As you can see, we havesquared the modulus and doubled the argument.

Continuing to multiply a complex number by itself, we get aresult known as de Moivres Theorem:

Example: Simplify .

The modulus of 1 + i is , and we can chooseA= 45 degrees.

Thus, =

=

= 16 cis(360)

cisnnz z nA

18

i

2

8

2 cis(845)

18 i

42


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Simplify

After computing by hand as in the previous

example, check your answer by calculator.

3

3 i

43


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Roots of complex numbers:

We now use de Moivres Theorem to find roots:suppose that and we would like to

solve forwin terms ofz. Writing the polar forms

of both sides of this equation, we have

or . Equating the moduli of each side, we have

,

hence .

Since the arguments of each side are equal (or,

differ by a multiple of 360 degrees), there is an

integerkso that

, hence .

w zn 0

w Bcis cisn

z A w nB z Ancis cis

w zn w zn

nB A k 360 BA k

n

360

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We conclude that the polar form of the nth root of

zis

Since kcan be anyinteger, it may appear that

there are infinitely many roots. But since cosine

and sine have period equal to 360, we get

different roots only fork= 0, 1, 2, , n-1: when k

= n, we get the same cis value that we got fork

= 0; when k= n+1, we get the same cis value

that we got fork= 1, etc. We conclude that a

nonzero complex number has exactly nnth roots.

z zA k

nn n

cis

360

45


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Example: Find the three cube roots of z= -8.

For the modulus, we have . For theargument, we have A = 180, so that fork= 0, 1,

2, the three values of

are 60, 180, and 300. Thus, the three cube roots of -

8 are: 2 cis 60, 2 cis 180, and 2 cis 300, which aresimplified as:

The first of these three is the principal root, the onecorresponding to k= 0.

You can try to check these answers on yourcalculator, but dont expect to get all three answers.The TI-84 seems to give a real answer, is there isone. If none of the roots are real, the calculator gives

the principal root. Let us check this for the cube root- - ^ -

8 23

180 360

3

k

1 3 2 1 3 , ,

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Find the four 4th roots of z= -16, and check youanswer on a calculator. You should find that there

are no real roots, so the calculator displays the

principal root:

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Lesson 2 - Quiz

1. Let and . Sketch theparallelogram determined by Also

sketch the parallelogram determined by

.

2. Find the polar from of .

(a) 2 cis 120 (b) 2 cis 150 (c) 2 cis 210

3. Find the standard form of z = 4 cis 135.

(a) (b) (c)

4 Sk t h th t ti th t l


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4. Sketch the vectors representing the two complex

numbers

5 cis 45 and 3 cis 90. Sketch and write the

product in polar form.

(a) 15 cis 135 (b) 8 cis 135 (c) 15 cis

4050

5. Explain why

6. The following are the cube roots ofi. Which one

is principal?(a) (b) (c)

Answers: 2(c), 3(b), 4(a), 6(b)

Lesson 3: Applications


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Solving quadratic equations

Remember that the quadratic formula solves thequadratic equation "ax2 + bx+ c= 0" for the

values ofx(called zeros of the equation). They

called zeros because they are the values ofxthat

make y= 0, in the quadratic function y= ax2 +bx+ c.

Recall that a quadratic equation has two, one, or

no realzeros, depending of the sign of the

discriminant , which appears under the squareroot in the quadratic formula:x

b b ac

a

2 42

50

Lesson 3: Applications

Two zeros occur when the discriminant is positive


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Two zeros occur when the discriminant is positive.

One zero occurs when the discriminant is zero.

There are no realzeros when the discriminant isnegative.

However, there are zeros if we allow complex numbers.We illustrate by an example.

Example: Find the zeros of , where c= 0, 1, 2.

In simple cases such as this, it is more convenient tofind zeros by factoring than by use of the quadraticformula.

In the case c= 0, we have ,

so the zeros are 0 and 2, and these are thex-intercepts of the graph of the parabola .

In the case c= 1, we have , sox=1 is the only zero, and the graph of the parabola

has it vertex at (1, 0).

In the case c = 2, the quadratic formula gives no realroots, but two com lex roots 1 i. The ra h of the

x x c2 2 0

x x x x2 2 2 ( )

y x x 2

x x x2 22 1 1 ( )

y x x 2

2 1

y x x 2 2 2

51

Comparison of the roots of 3 quadratic


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p q

equations

Please

compare thethree equations

and the

solutions.

Why the thirdequation

doesnt have a

real solution?

Graphically, canyou understand

whether or not

an equation has

a complex root? 52


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For , find values ofcwhichillustrate each of the cases above.

The example above shows the existence of

polynomials with no real zeros. By contrast,

polynomials always have zeros if we allow

complex numbers:

x x c2

4 0

53

Fundamental Theorem of


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Fundamental Theorem of

Algebra:

54

A polynomial equation of degree at least one has at least one zero in the complex number system.

You may wonder why we should care about zeros being complex numbers. You may expect thatthey would provide no useful real world information and should be considered as extraneous.

However, Eulers formula

=

cos+

sin provides a link between the complexexponential function and the real trigonometric functions sine and cosine, which are used to modelperiodic behavior. Such behavior is found in AC (alternating current) circuits and in vibrating or

oscillating systems. These systems are governed by differential equations of the form +

+ = 0. Solutions to this equation are exponential functions of the form = , whereris a zero of the quadratic equation

2 + + = 0. (Those who remember their calculus can check this by plugging = into the differential equation.) The complex zeros of this quadratic equation give oscillating

solutions via Eulers formula = cos + sin . The value ofrprovides information on thefrequency and period of the oscillation. Thus, the complex zeros of a quadratic equation can

provide real world information in certain applications.

E l Fi d th f


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Example: Find the zeros of .

Using the quadratic formula, we get

=

=

Although the TI-84 calculator has a SOLVE

function, it can find only real zeros. Of course,

you can check if your zeros are correct by

substituting them back into the original equation.


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Find the zeros of , andcheck your answers by substitution.
http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/addition_and_subtraction_of_complex_numbers_.html


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Further applications:

Engineers also use complex numbers in analyzingstresses and strains on beams and in studying resonancephenomena in structures such as planes and bridges.Instead of just one differential equation, you may have asystem of many differential equations, which leads tomatrix analysis. The complex numbers come up when you

seek the eigenvalues and eigenvectors of a certain matrix.The eigenvalues are roots of a certain polynomial equationassociated with this matrix. The matrices may be quitelarge, perhaps 1000 by 1000, and the associatedpolynomials are of very high degree.

Complex numbers are also used in such fields as digitalsignal processing, digital image processing, quantummechanics, and fluid dynamics. In this last case, complexfunctions are used to describe two-dimensional flow, forexample, flow around a pipe.

In summar com lex numbers are an indis ensible tool of

Applets and Links


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Applets and Links Arithmetic Operations with Complex Numbers

http://www.walter-fendt.de/m14e/complnum.htm

Addition And Subtraction Of Complex Numbers

http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/addition_and_subtraction_of_complex_numbers_.html

Addition and Subtraction Of Complex Numbers

http://www.poisonapplet.com/complex_numbers/addition_and_subtraction/

Everything about Complex Numbers: http://www.poisonapplet.com/complex_numbers/

Anotherhttp://www.ies.co.jp/math/java/comp/index.html

http://www.poisonapplet.com/complex_numbers/

Multiplication And Division

http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/multiplication_and_division.html

De Moivre's Theorem

http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/de_moivre_s_theorem.html

Conjugate Roots Theorem and plotting equations

http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/conjugate_roots_theorem.html Quiz:

http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=na

Content with simple applets:

http://www.analyzemath.com/complex/complex_numbers.html with simple applets

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http://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://www.poisonapplet.com/complex_numbers/addition_and_subtraction/http://www.poisonapplet.com/complex_numbers/http://www.ies.co.jp/math/java/comp/index.htmlhttp://www.poisonapplet.com/complex_numbers/http://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/conjugate_roots_theorem.htmlhttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.analyzemath.com/complex/complex_numbers.htmlhttp://www.analyzemath.com/complex/complex_numbers.htmlhttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-827999-2&chapter=5&lesson=9&headerFile=10&state=nahttp://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/complex_numbers/conjugate_roots_theorem.htmlhttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://www.poisonapplet.com/complex_numbers/http://www.ies.co.jp/math/java/comp/index.htmlhttp://www.poisonapplet.com/complex_numbers/http://www.poisonapplet.com/complex_numbers/addition_and_subtraction/http://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exe


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Fractal trip : Download program and install it:

http://prdownloads.sourceforge.net/xaos/winxaos

31.pre3.zip

Julia and Mandelbrot Set Explorer

http://aleph0.clarku.edu/~djoyce/julia/explorer.html

Fractal picture galary:

http://www.angelfire.com/art2/fractals/bestpics.ht

m
http://www.angelfire.com/art2/fractals/bestpics.htmhttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://prdownloads.sourceforge.net/xaos/winxaos31.pre3.ziphttp://prdownloads.sourceforge.net/xaos/winxaos31.pre3.ziphttp://aleph0.clarku.edu/~djoyce/julia/explorer.htmlhttp://www.angelfire.com/art2/fractals/bestpics.htmhttp://www.angelfire.com/art2/fractals/bestpics.htmhttp://www.angelfire.com/art2/fractals/bestpics.htmhttp://www.angelfire.com/art2/fractals/bestpics.htmhttp://www.angelfire.com/art2/fractals/bestpics.htmhttp://www.angelfire.com/art2/fractals/bestpics.htmhttp://aleph0.clarku.edu/~djoyce/julia/explorer.htmlhttp://prdownloads.sourceforge.net/xaos/winxaos31.pre3.ziphttp://prdownloads.sourceforge.net/xaos/winxaos31.pre3.ziphttp://localhost/var/www/apps/conversion/tmp/scratch_7/fractalprogr/xaos/xaos.exehttp://www.angelfire.com/art2/fractals/bestpics.htm


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Fractals The mathematics behind fractals are incredibly interesting and captivating. You need to have a grasp on algebra and

some complex number background is preferable. We already described how fractals are created through applyingfunctions, but never explained any functions and how they work. In this section, we will describe the two most popular

fractal sets and how they work, the Julia set and the Mandelbrot set.

To understand fractals, you need to understand complex numbers. Complex numbers are a way to put two coordinates

(x,y) into one number with two parts. One is a real number, which is any regular number like 3, 8.5, or 12/45. The other

is an imaginary number, which is defined as the square root of a negative number, and is characterized by i (defined as

i^2=-1, therefore i=sqrt -1) times a coefficient.When you take a number and square it, it always becomes positive. So

how do you take the square root of a negative number? You can't, that's why it's called imaginary. So, complex

numbers are made up of a real number plus an imaginary number. Examples include (1+.343i), (pi+343.6i), and (0+3i).Complex numbers are used in fractals because the real number is used to represent the x coordinate, and the complex

number is used to represent the y coordinate. So, if the computer wanted to iterate (3,8), it would apply the function to

(3+8i). This way, the function is dealing with a number to which most of the mathematical properties such as the

associative and distributive laws can be applied, instead of a set of x and y coordinates. It is important to note that the

complex coordinates are not the same coordinates of the pixel they represent. Pixel coordinates are always from 0 to

the bounds of the screen, usually something like (786, 233). The range we use depends on the fractal, but it is usually

something like x: -3 to 3; y: -2 to 2. Therefore, to apply the function to a pixel, we divide the units into hundreds of tiny

segments, and the computers deal with the tiny fractions.

We must also set a limit of iteration on our fractal. Since the points inside the Mandelbrot set never leave the screen,

we will iterate our function forever if we wait for them to leave our circle. To get around this, we set a limit on the

number of times we will iterate it. If the point is still in our circle after that many iterations, we assume it is part of the

set. The more iterations we use, the more exact and detailed our image will be, but the longer it will take to generate.

When we have done this with every pixel, we have a fractal. Other equations than this one produce different fractals.

Mandelbrot sets are produced the same way as Julia sets, except that c is different for every point. When generating a

Mandelbrot set, c is equal to the point we are determining the color for. We start with 0, the origin. Then we square it

and add c. We square this new value and again add c. When this finally leaves the circle, or when we have reachedour iteration limit, we color the point at the complex coordinate c. Then we move to the next point. C is changed to that



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video

Video Annenberg

Watch until 4:33

module on complex numbers

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