Approximating the Negative Imaginary Root of f(z) = z^4 - 1 using Newton's Method

8/6/2019 Approximating the Negative Imaginary Root of f(z) = z^4 - 1 using Newton's Method

http://slidepdf.com/reader/full/approximating-the-negative-imaginary-root-of-fz-z4-1-using-newtons 1/37

Abhishek Kumar

Candidate Number: 0662-024

- 1 -

Approximating the Negative Imaginary Root of 4

( ) 1 f z z using Newton‟s Method

IB Extended Essay: Mathematics

Abhishek Kumar

Advisor: Mr. Daniel Davis


Word Count: 3952



Abhishek Kumar


- 2 -

Abstract

Newton‟s Method is a method of numerical approximation, generally used toapproximate the real roots of a function, which exist in the real Cartesian plane. However,

it can be applied to complex functions, particularly 1)( 4 z z f . This paper investigates

the use of Newton‟s Method to approximate the negative imaginary root of 1)( 4 z z f .

When the initial approximation is purely imaginary, Newton‟s Method behaves similarly

as to when it is applied to approximate real roots of a function with purely real initialapproximations. Each subsequent approximation to the negative imaginary root of f ( z) is

found by finding the zero of the line tangent to f ( z) at the initial approximation,0

z .

The geometry of Newton‟s Method is quite different when the initial approximationcontains both a real and imaginary part. With such an initial approximation, a three

dimensional view of Newton‟s Method is required to visualize its dynamics. Firstly, themodulus of f ( z) is taken, forming a solid shell around the function, while preserving itsroots. Secondly, each subsequent approximation to the root of f ( z) is sought by finding

the intersection between the line in the ( x, y) plane containing the gradient vector, 0 z f ,

of )( z f at the initial approximation of 0

z , and the plane tangent to the surface of )( z f

at0

z .

Studying attraction Basins known as Newton Basins, which label all the points thatconverge to a certain root a single color, provides further insights into the geometry

of 1)( 4 z z f . These insights include the nature of the surface of the level sets of

)( z f at different points, and the geometry of the tangent plane and the lines containing

the gradient vectors of )( z f .

Word Count: 268



Abhishek Kumar


- 3 -

Contents

Section Page

Introduction 1

Overview of Newton‟s Method 2

Overview of Complex Numbers 4

Geometry of f ( z) = z4-1 5

Newton‟s Method Adapted for The Complex Plane 8

Approximating the Negative Imaginary Root of f ( z) with Purely Imaginary

Initial Approximations

9

Approximating the Negative Imaginary Root of f ( z)=z4-1 with Complex

Initial Approximations

12

Altering the Geometry of f ( z) = z4-1 13

A Three-Dimensional View of Newton‟s Method 17

Newton Basins 23

Conclusion 28

Bibliography 29

Appendices 30Acknowledgements 34



Abhishek Kumar


- 4 -

Introduction

Newton‟s Method is commonly used as numerical method of approximation of the realroots of a function. However, not too often is Newton‟s Method seen applied to complexfunction in which it is used to approximate the imaginary roots of a function. The study

of complex analysis begs the question: how a method, generally applied to real functions,will converge, geometrically, to the imaginary roots of a complex function? This topic is

worth discussing because it provides new insights into the geometry of complex functions

and varied versions of Newton‟s Method, which combine both the concepts of calculusand analytic geometry. This paper seeks to answer this very question through exploring

the use of Newton‟s Method in approximating the negative imaginary root of the

function 4( ) 1 f z z .



Abhishek Kumar


- 5 -

Overview of Newton‟s Method

Newton‟s Method works as follows: a value, which is reasonably close to the root of thegiven function, is chosen. Then the function is replaced by a tangent line, which is found

with the derivative of the given function, and the root of that tangent line is computed.

The root of this tangent is typically a better approximation to the function‟s root, and this

method is iterated to reach a very precise approximation of the function‟s root.

Suppose : [ , ] f a b is a differentiable function, which is defined on the interval [ , ]a b

with values in the real numbers, . The iteration process is started with the arbitrary

valuen

x (if the value is closer to the root of the function, it will take less iterations to

gain a precise approximation) and then each subsequent iteration is taken, [1]:

The formula used for Newton‟s method derives the next approximation to aparticular zero of a function. In other words, it finds the xn+1 approximation from the nth

approximation, xn. In this graph of f(x), the point-slope form of the equation tangent to f(x)

at , ( )n n x f x is ( ) '( )( )n n n

y f x f x x x . Now, the value of xn+1 would be found by

finding out where this tangent line crosses the x-axis, and we can do this by setting y = 0:

[1]Wikipedia, Newton‟s Method

1n x n

x

, ( )n n x f x

( ) y f x

y

x

Change in „ x‟

Change in „ y‟

Root Sought



Abhishek Kumar


- 6 -

0 ( ) '( )( )

( ) '( ) '( )

'( ) '( ) ( )

( )

'( )

n n n

n n n n

n n n n

n

nn

f x f x x x

f x x f x x f x

x f x x f x f x

f x

x x f x

[2]

The value of x that we have derived here is the value of the next approximation, xn+1.

Therefore,

1

( )

'( )

nn n

n

f x x x

f x , which is Newton‟s method.

[2]Finney, Ross L. 222-223



Abhishek Kumar


- 7 -

Overview of Complex Numbers

Unlike real numbers, which are the field of numbers, complex numbers are the fieldof numbers of the form x iy , where x and y are real numbers and i is the imaginary unit

equal to the square root of -1, 1 , [3 ]. A single letter, z, is often used to denote a

complex number, z x iy . In this component notation, z x iy can be written ( , ) x y .

In other words, the complex number z x iy is identified by the two-dimensional

real vectors 2( , ) x y , and this correspondence is denoted by: ( , ) x iy x y .

Additionally, the field of complex numbers includes the field of real numbers as a

subfield, [4].

[3]Weisstein, Complex Number

[4]Wikipedia, Complex Number



Abhishek Kumar


- 8 -

Geometry of 4( ) 1 f z z

Visualizing the complex functions is perhaps the most challenging part of the endeavor.Real functions are quite easy to visualize, as they exist simply on a Cartesian plane,

comprising of an x-axis and a y-axis. For instance, the function 2( ) 1 f x x exists only

on the ( x, y) plane; like so:

-3 -2 -1 1 2

2

4

6

8

It is quite easy to visualize such a function and its roots, which we can see exist at 1 x

and 1 x . However, a complex function is much more difficult to visualize, as it does

not exist in the Cartesian plane. For instance, consider the function, 4( ) 1 f z z . This

function has four different roots. However, we cannot graph this function on a simpleCartesian plane to obtain both its real and imaginary roots, as the function being analyzedis not real, but is complex. In other words, the Cartesian plane comprises of an x-axis anda y-axis, which can be written as the x-axis and the f(x)-axis, respectively.

Therefore, , , ( ) x y x f x , [5]. But the function that is being dealt with is a complex

function, which cannot lie on an , ( ) x f x plane, as its function is f ( z). Therefore, a new

plane for such a function must be brought into use.

From the function alone, it can be understood that one of the axis of this plane will be the

f ( z)-axis. For the other axis, recall section „Overview of Complex Numbers,‟ where it is

explained that a complex number consists of two parts; the complex number z consists of

a real and an imaginary part (real: x; imaginary: iy), which form the complex

number z x iy . Therefore, behaving like a real function, the real root, x, of f ( z) will lie

where the graph of f ( z) intersects an x-axis; where f ( z) = 0. So, for the graph of f ( z)=0 and

there exist a real root, such an x-axis must exist. Thus far, the following picture of thisplane exists:

[5]Gullberg 337-338



Abhishek Kumar


- 9 -

A plane displaying the real axis, x, of a complex number and the axis, f(z).

Thus far, this plane only yields the real root of a complex function. So, to obtain a valuefor the imaginary root, iy, of the complex number, there must be another axis added.

Following the same logic as that of adding the real axis, x-axis, to our graph, it can beassumed that imaginary root of the complex function will exist where the graph of f ( z)

intersects the “imaginary-axis.” In other words, there exists an imaginary root for f ( z)

where f ( z) = 0. Continuing with this logic, to represent the complete graph of a complexfunction, a third axis must be added; the y-axis, which represents the imaginary part of

the complex function:

Complex Plane – Representing three axis: the real part of a complex number, x-axis; the imaginary part of

a complex number, y-axis; and the axis, f(z).

x

f ( z)

y

x

f ( z)



Abhishek Kumar


- 10 -

A model capable of displaying the complete graph of a complex function has now been

derived, [6] [7]. Since 4( ) 1 f z z has two real roots and two imaginary roots in perfect

symmetry to each other, f ( z) will be graphed like so:

Figure-1

The function, f(z), graphed on the Complex Plan. For easy visualization, the curve intersecting the real

axis, x, is colored red, similar to the color of the x-axis itself; and the curve intersecting the complex axis,

y, is colored blue, similar to the y-axis itself.

This model encompasses the complete graph of f ( z). Both the real and the imaginary rootscan be clearly seen. Additionally, in the plane, the orthogonally intersecting axis, x and y,

combine to form a plane, which contains all the real numbers: x iy . In other words, the

x-axis of this plane is the Re( z) axis of an Argand Diagram and the y-axis is the Im( z) of

an Argand Diagram. Therefore, the x-axis will contain all the real values and the y-axis

will contain all the imaginary values. So, basically, an Argand Diagram has beencombined with an orthogonally intersecting axis, f ( z).

Combining these three axis states that when considering only real values, the imaginaryvalues will be considered zero; and when considering only imaginary values, real values

will be zero. Also, combining the three axis also allows f ( z) to be written as f ( x, y), as thevertical axis is a function of both the x and the y axis.

[6]Visualizing Functions of a Complex Variable

[7]Bronwell 339-383

x

f ( z)

y

f ( z)



Abhishek Kumar


- 11 -

Newton‟s Method‟s Adapted to the Complex Plane

The root of any functions exists where that function equals to zero. Say, the function, f ( x)is to be taken. The root of f ( x) will exist where f ( x) = 0. This exact same logic can be

applied to roots of complex functions as well. Say, the complex function, f ( z), is taken.

This function has complex roots, and these roots of f ( z) will exist where f ( z) = 0.Therefore, roots of both real and complex functions exist where the functions equal zero,

[8]. Considering this commonality between the real and complex functions, implies that

the root of a complex function can be approximated by Newton‟s method as well.

However, the complex Newton‟s method would have to be written differently, and havedifferent rules:

The real Newton‟s method is written like so:

1

( )

'( )

nn n

n

f x x x

f x

And has the rule that x . However, the complex Newton‟s Method will be written like

so:

1

( )

'( )

nn n

n

f z z z

f z

And has the rule that z , where0

z , our first approximation, is “sufficiently” close to

the root sought, [9], [see section, „Newton Basins‟].

[8]Morris 623-669

[9]Shepperd



Abhishek Kumar


- 12 -

Approximating the Negative Imaginary Root of f ( z)

with Purely Imaginary Initial Approximations

When a purely imaginary value is plugged in for the complex number, z x iy , the

result is a purely imaginary number. Therefore, a visual depiction of the graph of f ( z)when abiding by the geometry derived in the section, „Geometry of 4( ) 1 f z z ‟ and

considering only purely imaginary values would look like so:

Graph of f(z)=z4-1, considering only purely imaginary values.

Now, considering the above graph, the complex Newton Method [refer to „Newton‟sMethod Adapted for the Complex Plane‟] can be used to approximate the negative

imaginary root of f ( z):

1

( )

'( )

nn n

n

f z z z

f z

The first step would be to choose an initial approximation to the negative imaginary root

of f ( z). Because the closest whole number to the left of the root seems to be -2i, the first

approximation must be z=-2i; in other terms, 0 2 z i . With all the necessary information

obtained, the negative imaginary root shall be approximated with Newton‟s Method:

01 0

0

( )

'( )

f z z z f z

Differentiating the proper terms,

4

01 0

4

0

0

1

1

z z z

d z

dz

f ( z)

z

Note: f(z) here is purely

imaginary.



Abhishek Kumar


- 13 -

4

01 0 3

0

1

4

z z z

z

Substituting -2i for0

z we obtain,

4

1 3

1

2 12

4 2

1.531

i z i

i

z i

Iterating further,

4

2 3

2

1.531 11.531

4 1.531

1.218

i z i

i

z i

Without showing work,

3

4

5

6

1.052

1.004

1.000

1.000

z i

z i

z i

z i

The above iterations can also be seen occurring graphically:

The black curve represents the function f(z), and the red lines are tangent lines representing each iteration.

Since the fifth iteration yielded no change in the fourth significant figure, it will be

presumed that the negative imaginary root of f ( z) is z i . Newton‟s Method has been

z

f ( z)

1 1.531 z i 2 1.218 z i



Abhishek Kumar


- 14 -

successfully employed to approximate the negative imaginary root of f ( z). Therefore, we

can see that Newton‟s Method can be used to approximate the real and imaginary roots of

the function 4( ) 1 f z z if the approximations are taken only in the form of purely real

or purely completely imaginary. However, so far, only a set case of purely imaginary or

purely real values has been investigated. Cases dealing with initial approximations of the

complex form x iy , which are neither purely real nor purely imaginary, have not yetbeen considered.



Abhishek Kumar


- 15 -

Approximating the Negative Imaginary Root of 4( ) 1 f z z with Complex Initial Approximations

In the previous section, approximating the negative imaginary root of 4( ) 1 f z z ,

provided the initial approximation was purely imaginary, was discussed. However, what

would be the geometry of Newton‟s Method if the initial approximation contained both a

real and an imaginary part? The next two sections are aimed at discussing that veryphenomenon.



Abhishek Kumar


- 16 -

Altering the Geometry of 4( ) 1 f z z

If the initial approximation is complex, containing both real and imaginary parts, thegeometry of f ( z) needs to be altered to apply Newton‟s Method [refer to „A Three-

Dimensional View of Newton‟s Method‟].

Upon viewing the graph of f ( z) [see Figure-1 in „Geometry of 4( ) 1 f z z ‟ section], it

seems as if it is forming a shell:

From the above graph, it may seem as if there is a paraboloid, [10], connecting the roots of the function to form a solid shell, but that is not the case. The function is disjointed; someof its values are below the axis and some are above the axis. For instance, it is clear from

the graph of f ( z) that there are no z values for which the function has a value smaller than

f ( z)=-1. However, if we take the point 1.15 1.15 z i , f ( z) yields the value -8. It is quite

obvious from the graph of f ( z) that the function does not exist at f ( z)=-8. This shows that

some of the points, not visible, are disjointed from the graph of f ( z). Another example of the disjointedness of the function is that the graph clearly exists only in either the plane,

, ( ) x f z or , ( ) y f z , but not anywhere else; looking at Figure-1 from an aerial view

may assist in visualizing such an abstract concept:

[10]Wikipedia, Paraboloid

x

f ( z)

y



Abhishek Kumar


- 17 -

An aerial view of the (x, y, f(z)) plane [Figure-1]. The real roots of f(z) are presented as blue dots on the

real axis, x; and the imaginary roots of f(z) are presented as red dots on the imaginary axis, y.

The real portion of the function f ( z) passes vertically upwards through the blue points and

the imaginary portion of f ( z) passes vertically upwards through the red points. Therefore,

it is evident that ( ) 0 f z does not exist at, say,1 1

2 2 z i [shown below], which is

on the circle of intersection of the apparent paraboloid.

The green dot denotes the point 1/ 2 1/ 2 z i

Substituting1 1

2 2 z i into f ( z) should yields the value f ( z) = 0 if the paraboloid

exists, but the value it yields is f ( z) = -2. In fact, anything on the diagonals, y x

or y x i.e. 1 , 2 2 , etc.i i , results in ( ) < 0 f z . Therefore, it can be understood that

y

x

y

x



Abhishek Kumar


- 18 -

the paraboloid does not exist, proving that f ( z) is disjointed. Therefore, to correctly

implement Newton‟s Method with the initial approximation as a value other than one onthe x or the y axis [see section, „A Three-Dimensional View of Newton‟s Method‟], all

the disjointed points must be brought together; all must be collected within the positive

section of the axis, f ( z). The simplest way to assure that all points f ( z) values are positive

is by taking the modulus of f ( z), like so:4

( ) : 1 f z z , [11] [12]. Taking the modulus of f ( z) will assure that all the points are positive, and as an added bonus, it will create a

function with a solid shell, enabling us to easily implement Newton‟s Method to

approximate the roots:

Screenshots taken from a TI-89 calculator of the function f (z) and the function | f (z)|, respectively. See

„Appendix-[1]’ for the programming inputted into the TI-89.

[11]Richards 319

[12]Weisstein, Complex Modulus

Converting from graph of f (z) to

graph of | f (z)|



Abhishek Kumar


- 19 -

From the above graph, it can be noticed that the graph of | f (z)| has a solid shell. Thisparticular geometry will assist understanding the geometry of Newton‟s Method greatly.



Abhishek Kumar


- 20 -

A Three-Dimensional View of Newton‟s Method

When dealing with a two-dimensional space, the geometric interpretation of Newton‟s

Method is well known: the next iterate1n

x is the root of the line

( ) '( )( )n n n y f x f x x x tangent to the graph of f ( x) at )(, nn x f x . However, for a

complex function, this process is carried out not by tangent lines, rather tangent planes.

In the previous section, a solid shelled version of f ( z) was derived. This solid shell will

enable planes to be tangent against that very shell. To visual this a bit clearer, the graph

and level sets of | f ( z)| are given below:

The function | f ( z) | graphed on the (x, y, f ( z)) plane with its level sets visible; showing the two zeros, z=i

and z=-i.

Figure-1

Given the above graph, a visualization of the tangent plane can be taken, along with other

components: In the real case, each subsequent approximation, 1n x , is found by finding

the root of the line tangent to the function atn

x . In the complex case, however, each

subsequent approximation,1n

z , is found by finding the intersection between the line

f ( z)

| f ( z) |

y

x

i i



Abhishek Kumar


- 21 -

containing the gradient vector,n f z , and the plane tangent to the function at

n z . The

diagram below puts all of these planes and vectors into perspective, [13]:

The plane tangent to | f ( z)| at zn is labeled as T n , the gradient vector orthogonal to the tangent plane and the

graph of | f ( z)| at zn is represented by the arrow, f zn .

Figure-2

In the above illustration, it is noticeable that the subsequent approximation to the sought

root is the intersection between the line containing the gradient vector n f z and the

plane, T n. To see a working example, this method of approximating the root of a complex

function can be applied to the function 4( ) 1 f z z .

As stated in the previous section, we must use the modulus of the f ( z) in order to applythe complex Newton‟s Method. Applying the modulus will create a solid shell around the

graph of the function, allowing planes to be tangent to its surface. To begin applying

Newton‟s Method, an initial point must first be denoted: 0.123 0.68n z i . However,

the imaginary number, i, is accounted for in the function 4( ) 1 f z z , as z x iy .

[13]Yau and Ben-Israel

f ( z)

| f ( z)|

y

x zn

zn+1

| f ( zn)|T n

n f z



Abhishek Kumar


- 22 -

Therefore, the initial point will be taken as two separate real numbers, ( , ) x y [refer to

section, „Overview of Complex Numbers‟]. Thus, the initial approximation

is ( 0.123, 0.68) . With the initial approximation at hand, the tangent plane can now be

found:

The equation of any plane tangent to a function is given by the following formula, [14]:

0 0 0 0 0 0 0 0, , , f f

z f x y x y x x x y y y x y

Finding values for all the variables in the equation of a tangent plane, while considering

the function 4( ) 1 f z z :

Finding 0 0, f x y :

4

0 0 0 0

4

0 0

, 1

0.123, 0.68 0.123 0.68 1

, 0.8413 4 S.F.

f x y x iy

f i

f x y

Finding 0 0, f

x y x

:

6 4 2 2 4 2 4

28 6 2 4 4 2 2 4 4

4 3 3 1 3,

4 2 3 1 4 3 1

0.123, 0.68 0.8660 4S.F. See Appendix for Working

x x x y x y y y f x y

x x x y x y x y y y

f

x

Finding 0 0, f

x y y

:

6 4 2 2 4 2 4

28 6 2 4 4 2 2 4 4

4 3 3 1 3,

4 2 3 1 4 3 1

0.123, 0.68 0.9961 4 S.F.

y y y x y x x x f x y

y y y x y x y x x x

f

y

[14]Mardsen and Tromba 124-151



Abhishek Kumar


- 23 -

Substituting values into the equation of the tangent plane:

0.8413 0.866 0.123 0.9961 0.68 z x y

Since we‟re considering only the part of the tangent plane that lies in the ( , ) x y plane of the ( , , ( )) x y f z plane, the value for z in the above equation will be zero. Thus, giving the

following equation for the plane tangent to | f ( z)| at | f ( z0)|:

0 0.8413 0.866 0.123 0.9961 0.68 x y (0.1)

The next step is to find the gradient vector through the following equation:

0 0

0 0

,

,

f x y

x

f f x y

y

Since these values have already been found in the process carried out above, the gradient

vector of | f ( z)| at | f ( z0)| is:

0.866

0.9961 f

The next step in finding the value of 1

z (the next approximation to the root) is creating a

line, which contains the gradient vector found above:

0

0

x x f

y y

0.123 0.866

0.680 0.9961

x

y

(0.2)

To find the point of intersection between the line containing the gradient vector (0.2) andthe tangent plane (0.1), the value of λ must be sought by substituting the x and y values of

the lines into the plane:

0.123 0.866

0.68 0.9961

x

y



Abhishek Kumar


- 24 -

Substituting these values of x and y into equation of the tangent plane (0.1) and solving

for λ :

0.8413 0.866 0.123 0.866 0.123 0.9961 0.68 0.9961 0.68 0

0.4687 4S.F.

Substituting value of λ into the equation of the line will yield the point of intersection:

0.123 0.866

0.680 0.9961

0.123 0.8660.4687

0.680 0.9961

0.2952 1.161

x

y

x

y

x and y

These values can now be substituted into the complex number z x iy to obtain the

second approximation to the negative imaginary root of | f ( z)|:

1 0.2952 1.161 4S.F.

z x iy

z i

Continuing this entire process of iteration, Newton‟s Method will converge like so:

2

3

4

5 5

0.1226 0.9775 4 S.F.

0.0037 0.9765 4 S.F.0.0003 1.001 4S.F.

0 4 S.F.

z i

z i z i

z i z i

Through successful implementation of analytic geometry concepts, it has been shown that

Newton‟s Method can converge to the sought root when the initial point is other than onesolely on the real axis or imaginary axis. Furthermore, another interesting point that

arises is that Newton‟s Method, a two-dimensional method, yields the same

approximations as that of the abovementioned three-dimensional method:4

0

1 0 30

1

4

z

z z z

4

1 3

1

0.123 0.680 10.123 0.680

4 0.123 0.680

0.2952 1.161

i z i

i

z i



Abhishek Kumar


- 25 -

4

2 3

2

4

3 3

3

4

4 3

4

0.2952 1.161 10.2952 1.161

4 0.2952 1.161

0.1226 0.9775

0.1226 0.9775 10.1226 0.9775

4 0.1226 0.9775

0.0037 0.9765

0.0037 0.9765 10.0037 0.9765

4 0.0037 0.9765

0.0

i z i

i

z i

i z i

i

z i

i z i

i

z

4

5 3

5 5

003 1.001

0.0003 1.001 10.0003 1.001

4 0.0003 1.001

0

i

i z i

i

z i z i

Therefore, though the original method is intended for two-dimensional use, it yields thesame approximations as those of the three-dimensional adaptation of the method.



Abhishek Kumar


- 26 -

Newton Basins

Newton‟s Method has been seen converging the desired negative imaginary root of f ( z) in

the previous sections. However, the question that arises now is that which points will

converge to the root sought and which will not. A certain initial guess may approach one

root, and another guess may approach a different root. If we color the complex plane in amanner that designates a specific color to all the points that converge to a certain root, we

obtain „Newton Basins‟ (or domain of attraction), [15]. So, if a specific color basin

assigned to the rooti

, all the points within that color basin will eventually converge

toi

, [16]. Through the use of Mathematica 5, an image showing these Newton Basins has

been obtained by repeated iterations of the Newton‟s method applied to f ( z):

(See ‘Appendix-[3]’ for Mathematica Programming –

1)

(Figure-1)

Carried out just as stated above, the Newton basins have been created; each color

represents points that will converge to certain root. These Newton Basins will give some

insight into the dynamics of Newton‟s method. For instance, if an initial guess is to be

taken from the purple regions, Newton‟s method would converge to the imaginary

root, z i . This means that tangent planes are taken against the shell of the modulus of

f ( z) until the desired root is reached. To prove this statement, we will go through a few

examples.

Let us take our initial guess as the same one in the previous section, 0 0.123 0.680 z i ,

which is located in the purple basin presented in the figure below:

[15]Garcia

[16]Joyce

Root sought, z = -i.



Abhishek Kumar


- 27 -

Domain of Image: 0.353 ≥ Re( z) ≥ 0.313

Range of Image: -0.708i ≥ Im( z) ≥ -0.0417i

(Figure-2)

Beginning the iterating process,

4

01 0 4 1

0

1 z z z

nz

4

1 3

1

4

2 3

2

4

3 3

3

4

0.123 0.680 10.123 0.680

4 0.123 0.680

0.295 1.161

0.295 1.161 10.295 1.161

4 0.295 1.161

0.123 0.977

0.123 0.977 10.123 0.977

4 0.123 0.977

0.004 0.976

0.004

i z i

i

z i

i z i

i

z i

i z i

i

z i

z

4

3

4 4

0.004 0.976 10.976

4 0.004 0.976

0 (3 . .)

ii

i

z i z i S F

Picking a point in the same basin as the one containing the imaginary root z = -i has

yielded that very root when using Newton‟s Method. Geometrically, this shows that



Abhishek Kumar


- 28 -

Newton‟s Method converged exactly how it has been depicted in figure-1 of the previous

section. For a second example, we will choose a point in a purple basin that is surrounded

by different basins, making it far more difficult for Newton‟s method to reach thedesignated root. We will take our initial guess in a very small purple basin located in the

top left-hand corner of the complex graph of f ( z), as pointed out by the arrow in figure-3:

(Figure-3)

The process of zooming into desired purple basin is presented below in sequential order:

(Figure-4)



Abhishek Kumar


- 29 -

We will choose the point 0.6314 0.85133 z i from the indicated basin. The initial

guess contains more than three significant figures because this purple basin is so smallthat it requires the highest amount of precision. Now that we have an initial estimate, we

will begin iterating in hopes of finally reaching the root, z = -i.

0

4

0

1 0 3

0

4

1 3

1

4

2 3

2

8 7

,

0.6314 0.85133

,

1

0.6314 0.85133 10.6314 0.85133

4 0.6314 0.85133

0.276 0.586

0.276 0.586 10.276 0.5864 0.276 0.586

0.764 0.637

.

..

...

0.0

Given

z i

And

z z z

nz

i z i

i

z i

i z ii

z i

z z

4

3

8 8

0.017 0.987 117 0.987

4 0.017 0.987

0

ii

i

z i z i

Newton‟s Method has once again yielded the root specified by the color of the basin; the

root, z = -i, which is in a purple basin, has been located using Newton‟s method with theinitial guess also being a point in a purple basin. Though the method converged to the

designated root, it took more iterations to do so because the basin chosen was surrounded

by many other basins of different colors. Geometrically, this shows that the surface of the

shell around the modulus of f ( z) at0

z was very steep, as the distance of the step between

the first and second approximation was small. However, the surface at1

z quite flat, as it

produced a huge step, yielding the next approximation as 0.764+0.637i. We can also

presume that the tangent plane taken in the first approximation was vertically quite small,whereas the one in the second iteration was quite long, as was the gradient vector.

Another scenario can also take place in which Newton‟s Method doesn‟t converge to anyroot. An initial approximation that will yield such a result exists in the black areas of

figure-1. For instance, if the point 1+i is taken, the following results will follow:



Abhishek Kumar


- 30 -

4

01 0 4 1

0

1 z z z

nz

4

1 3

1

4

2 3

2

4

3 3

3

4

4

1 11

4 1

0.688 0.688 (3 . .)

0.688 0.688 10.688 0.688

4 0.688 0.688

0.323 0.323

0.323 0.323 10.323 0.323

4 0.323 0.323

1.607 1.607

1.607 1.6071.607 1.607

i z i

i

z i s f

i z i

i

z i

i z i

i

z i

i z i

31

4 1.607 1.607

.

..

...

Re[ ] Im[ ]n

i

z z z z

From the above calculations, it is evident that a point chose on the black areas of figure-1

will not converge to any root; more specifically, a point taken on the line Re[ ] Im[ ] z z

causes Newton‟s Method to oscillate back and forth on the same line. This shows thatshell of ( ) f z at the black Basins is so that it prevents the next approximation from

landing in any colored basin; thus preventing any convergence.



Abhishek Kumar


- 31 -

Conclusion

Whether the initial approximation is purely imaginary or has both a real and an imaginary

component, Newton‟s Method can be used to approximate the negative imaginary root of

4( ) 1 f z z . When the initial approximation is purely imaginary, Newton‟s Methodtakes on the same form, geometrically, as when applied to approximate real roots of afunction. On the other hand, a three dimensional form of Newton‟s Method is required tovisualize its dynamics against f ( z), and surprisingly, the results of this three-dimensional

variation are similar to those of the regular two-dimensional version of the method.

However, this discussion is not broad enough to explore how Newton‟s Method would

behave for a class of 1)( n z z f functions, where 5n . Such functions may produce

completely different visuals of Newton Basins, causing Newton‟s Method, geometrically,in both three and two dimensions when used to approximate the function‟s roots.Furthermore, another question that is left unanswered by this discussion is how Newton‟sMethod will behave geometrically when applied to complex functions not in the form

of 1)( n z z f . For instance, complex functions such as )()cos()( 6 z z z f . Therefore,

though this essay addresses numerous methods, which could be used to approximate the

negative imaginary root of 4( ) 1 f z z , it still leaves room for much further research in

this interesting topic.



Abhishek Kumar


- 32 -

Bibliography

[1] "Newton's Method." Wikipedia. 28 Feb. 2005. 2 Mar. 2005 <http://en.wikipedia.org/wiki/

Newton%27s_method>.

[2] Finney, Ross L., et al. "Linearization and Newton's Method." Calculus: Graphical,Numerical, Algebraic. Upper Saddle River, New Jersey: Pearson Prentice Hall, 2003.

222-223.

[3] Weisstein, Eric W. "Complex Number." MathWorld. Wolfram. 20 Jan. 2005

<http://mathworld.wolfram.com/ComplexNumber.html>.

[4] "Complex Number." Wikipedia. 25 Feb. 2005. 2 Mar. 2005

<http://en.wikipedia.org/wiki/Complex_number>.

[5] Gullberg, Jan. Mathematics: From the Birth of Numbers. New York, New York: W.W.

Norton & Company Inc., 1997.

[6] "Visualizing Functions of a Complex Variable." Pacific Tech. 2002. 18 Jan. 2005<http://www.pacifict.com/ComplexFunctions.html>.

[7] Bronwell, Arthur. Advanced Mathematics in Physics and Engineering. 1st ed. New York:

McGraw-Hill, 1953.

[8] Kline, Morris. Mathematical Thought from Ancient to Modern Times. Vol. 2. New York:Oxford University Press, 1990.

[9] Shepperd, Michael. "Newton Basins." Mathematics Resources. 21 Feb. 2005 <http://michaelshepperd.tripod.com/calcnewtoncomplex/>.

[10] "Paraboloid." Wikipedia. 1 Mar. 2005. 2 Mar. 2005<http://en.wikipedia.org/wiki/Paraboloid>.

[11] Richards, Paul I. Manual of Mathematical Physics. London; New York: Pergamon Press,

1959.

[12] Weisstein, Eric W. "Complex Modulus." Mathworld. 2 Feb. 2005

<http://mathworld.wolfram.com/ComplexModulus.html>.

[13] Yau, Lily, and Adi Ben-Israel. "The Newton and Halley Methods for Complex Roots." The

American Mathematical Monthly 105 (1998).

[14] Marsden, Jerrold E., and Anthony J. Tromba. "Differentiation." Vector Calculus. 3rd ed.

New York, New York: W.H. Freeman and Company, 1988. 124-154.

[15] Garcia, Fransisco, et al. "Coloring Dynamical Systems in the Complex Plane." Unpublished

essay.

[16] Joyce, David E. "Newton Basins: Newton's Method." Newton Basins. Aug. 1994. 22 Feb.

2005 <http://aleph0.clarku.edu/~djoyce/newton/method.html>.

http://www.questia.com/PM.qst?a=o&d=77293775















Abhishek Kumar


- 33 -

Appendix

Appendix-[1]:

TI-89 Programming:

i) Switch to 3D graphing by pressing [MODE] and switching ‘Graph’ to ‘3D’

ii) Press [diamond ] and input the following command in ‘ y1’: 1)( 4 iy x .

iii) Press [diamond ] and [F3].

Appendix-[2]:

Differentiating 1)(),(

4

iy x y x f with respect to x:

Expanding the function,

1464),(

1)(),(

432234

4

y xiy y xiy x x y x f

iy x y x f

Separating the real and imaginary parts,

334224

4416),(xiyiy x y y x x y x f

Simplifying the modulus of the function,

23324224 4416),( xy y x y y x x y x f

Without the imaginary unit, i, finding ),( y x x

f

becomes much easier,

Using chain rule,

2 24 2 2 4 3 3

2 24 2 2 4 3 3

16 1 4 4

2 6 1 4 4

d x x y y x y xy

dx x x y y x y xy



Abhishek Kumar


- 34 -

2 23 3 4 2 2 4

2 24 2 2 4 3 3

4 4 6 1

2 6 1 4 4

d d x y xy x x y y

dx dx

x x y y x y xy

Using the chain rule once again,

23 3 3 3 4 2 2 4

2 24 2 2 4 3 3

2 4 4 4 4 6 1

2 6 1 4 4

d d x y xy x y xy x x y y

dx dx

x x y y x y xy

22 3 3 3 4 2 2 4

2 24 2 2 4 3 3

2 12 4 4 4 6 1

2 6 1 4 4

d x y y x y xy x x y y

dx

x x y y x y xy

Using the chain rule yet again,

2 3 3 3 4 2 2 4 4 2 2 4

2 24 2 2 4 3 3

2 12 4 4 4 2 6 1 6 1

2 6 1 4 4

d x y y x y xy x x y y x x y y

dx

x x y y x y xy

2 3 3 3 3 2 4 2 2 4

2 24 2 2 4 3 3

2 12 4 4 4 2 4 12 6 1

2 6 1 4 4

x y y x y xy x xy x x y y

x x y y x y xy

Simplifying,

6 4 2 2 4 2 4

2 24 2 2 4 3 3

4 3 3 1 3

6 1 4 4

x x x y x y y y

x x y y x y xy

Using Binomial Expansion for the terms in the denominator,

6 4 2 2 4 2 4

28 6 2 4 4 2 2 4 4

4 3 3 1 3

4 2 3 1 4 3 1

x x x y x y y y

x x y x y x y y y



Abhishek Kumar


- 35 -

Appendix-[3]:

“Mathematica” Programming, in sequential order, behind Figure-1 in „Newton Basin‟section:



Abhishek Kumar


- 36 -



Abhishek Kumar


Acknowledgements

There are five people to whom I‟d like to pay my gratitude. I would like to begin bypaying my respects to a very special friend of mine, Daphne Tenne, because she was the

reason I became interested in mathematics. Once a lost young man, I gained my footingbecause of her beautiful words that embodied her unbelievable passion for mathematics.Though she has moved away, she still remains in my heart, inspiring me to achieve ever

higher and pursuing my dreams.

Soon after came one of my greatest mentors, Mr. Joseph Reing. I had no one to sharemy flourishing interest in mathematics during its early stages, except for Mr. Reing. He

listened to my everyday intrigues with the strange, abstract concepts of mathematics, and

he would carry on such discussions for hours on end. When most thought a high school

student would be incapable of writing an extended essay in mathematics, Mr. Reingbelieved in me, staying firm to the belief that if I pursued this assignment with a true

passion, I‟d be able to complete such a monumental task.

My father once had a son who was lost in his academic life, and even through that,

my father never lost hope in me. Throughout my entire life, he has believed in me

without a shadow of a doubt. He was one of the main reasons I decided to pursue amathematical topic as my Extended Essay. My father, with his never-failing support,

encouraged me to pursue a task, which at first seemed daunting. Because of his

unbelievable faith in me, I would like to thank him for being responsible for the birth of

this paper.

Moving onto my teachers, I‟d like to thank Mr. Brewster Campbell for believing in

me and my mathematical abilities to let me join his IB Mathematics HL class, whichserved as a great learning sources for me and allowed me to truly explore my newly-born

love of mathematics.

I must pay the highest gratitude to my advisor, Mr. Daniel Davis, who worked withme in each step of the way, helping me grasp newer concepts – concepts I never thought

I‟d learn in high school. He was an amazing help to me. I have taken two of his classes,

and they both blew me away because of how interesting he made them, and so workingwith him one-on-one was unbelievable. Therefore, for his dedication, and for his

unbelievable persistence and breadth of knowledge, I offer Mr. Davis the highest

gratitude.

Approximating the Negative Imaginary Root of f(z) = z^4 - 1 using Newton's Method

Documents