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IB MATHS HL POTFOLIO TYPE 1

Patterns in Complex Numbers

An analytical paper on the roots of a complex numbers and its geometry

1i

Syed Tousif Ahmed Candidate Session Number: 006644-009

School Code: 006644

Session: May 2013

Table of Contents 1. Introduction .......................................................................................... 2

2. Polynomials over the Complex Field .................................................... 3

3. Solutions to equations of the form , where zn is a real number ...................................................................................................... 9

4. Geometry of the nth roots of the equation zn – 1=0 ............................ 21

5. Solutions to equations of the form , where zn is not a real number .................................................................................................... 31

6. Generalizations of the results for zn = x+iy ........................................ 39

7. Conclusion ........................................................................................... 43

Syed Tousif Ahmed Candidate Session Number: 006644-009 2

1. Introduction Solving equations and finding answers to those equations has always been and will be a mathematician's past time; as it was for al-Khwarizmi, Cardano, Bombelli, Descartes and many other mathematical greats. Only a mathematician can realize the divine feeling of solving and resolving an equation and giving meaning to it. This paper analyzes the nth root of a complex number and also relates to the geometry of complex numbers. It identifies a conjecture relating to regular polygons and discusses how the study of complex numbers raises this conjecture. Lastly, the paper describes some application of the roots of a complex numbers in several areas of study.


2. Polynomials over the Complex Field

Before we can move onto the analysis of the nth root of a complex number, it is essential to note down some notations used in this paper, some definitions required to understand the mathematical language and some theorems which will be used in explicating the paper. However, this paper assumes that the reader is familiar with the manipulation of complex numbers. The angle measure in this paper is in radians.

2.1 Definitions 2.1.1 Polynomials: A polynomial function, P(x), is an algebraic expression that takes the form

1 2 11 2 1 0( ) ... , 0n n n

n n n nP x a x a x a x a x a a

Where the coefficients an, an-1, an-2, …, a1, a0 are real numbers and the power, n, n-1, n-2, … are inte gers. The degree of a polynomial, degP(x), is the highest power of x in the expression.

2.1.2 Complex Numbers: The set of complex numbers is denoted by

{ : ,C z z x iy where 2, , 1}x y R i

The real part of z, denoted by Re(z) is x. That is, Re(z) = x. The imaginary part of z, denoted by Im(z) is y. That is, Im(z) = y.


2.1.3 Polynomial over the complex field: When the coefficients, an, an-1, an-2, …, a1, a0, of the polynomial P(x) are complex numbers, the polynomial is a polynomial over the complex field.

2.2 Theorems 2.2.1 Remainder Theorem: For any polynomial P(x), the remainder when divided by (x – α) is P(α).

Proof: The degree of the remainder R(x) must be less than the degree of the divisor D(x). Therefore if D(x) has degree = 1, R(x) has degree = 0 and is therefore constant.

∴ if ( ) ( ) ( )P x D x Q x R and ( ) ( )D x x then

( ) ( ) ( )P x x Q x R (where R is a constant)

When x = α, ( ) ( ) ( )P Q x R

( )P R

i.e. the remainder on division of P(x) by (x – α) is P(α).

2.2.2 Factor Theorem: (x – α) is a factor of P(x) if and only if P(α) = 0.

Proof: By the remainder theorem, ( ) ( ) ( )P x x Q x R for all real x.

( )P R

But if ( ) 0P i.e. R = 0 then ( ) ( ) ( ) 0P x x Q x

( ) ( )x Q x

i.e. (x – α) is a factor of P(x).

2.2.3 Fundamental Theorem of Algebra: Every polynomial equation of the form P(z) = 0, z∈C, of degree n∈ ℚ+ has at least one complex root. A polynomial Pn(z), z∈C, of degree n∈ ℚ+, can be


expressed as the product of n linear factors and hence, produce exactly n solutions to the equation Pn(z) = 0.

Proof: By factor theorem, if P(z) has a solution z1 such that P(z1) = 0, then (z – z1) is a factor of P(z). Therefore, we have that

1 1( ) ( ) ( ) 0nP z z z P z where Pn-1(z) is itself a polynomial of degree n –

1. By applying fundamental theorem of algebra again, the equation Pn-

1(z) = 0 also has a solution, say z2, so that

1 2 2( ) ( )( ) ( ) 0nP z z z z z P z where Pn-2(z) is a polynomial of degree n

– 2. Continuing in this manner, after n applications we have that

1 2 2 0( ) ( )( )...( ) ( )P z z z z z z z P z where P0(z) is a constant. And so, we

find that there are n solutions, namely z1, z2, …, zn to the polynomial equation P(z) = 0.

2.2.4 Conjugate Root Theorem: The complex roots of a polynomial equation with real coefficients occur in conjugate pairs.

Proof: From the fundamental theorem of algebra we have that P(z) =

0. Then, it must be the case that ( ) 0 ( ) 0P z P z . Let 1 2

1 2 1 0( ) ...n n nn n nP z a z a z a z a z a

so that

1 11 1 0 1 1 0( ) ... ...n n n n

n n n nP z a z a z a z a a z a z a z a

1

1 1 0...n n

n na z a z a z a P z

And so, as ( ) 0 0P z P z . That is, z is also a solution.

2.2.5 De Moivre’s Theorem( (cos sin )) (cos sin ) (cos( ) sin( ))n n n nr i r i r n i n

( )nr cis n


Proof: (By Mathematical Induction)

Let P(n) be the proposition that ( ( )) ( ).n nrcis r cis n

For n=1, we have that L.H.S 1 1( ( )) ( ) (1 )rcis rcis r cis R.H.S

Therefore, P(n) is true for n=1.

Assume now that P(n) is true for n = k, that is, ( ( )) ( ).k krcis r cis k

Then, for n = k + 1, we have

1( ( )) ( ( )) ( ( ))k krcis rcis rcis

1

1

1

( )( ( ))( ) ( )( )(( 1) )

k

k

k

k

r cis k rcisr cis k cisr cis kr cis k

Therefore, we have that P(k+1) is true whenever P(k) is true. Therefore, as P(1) is true, by the Principle of Mathematical Induction, P(n) is true for n=1,2,3,…

For n being a negative integer:

Put n = -m, where m is a positive integer.

Then, 1 1 1 cos( ) sin( )

cos( ) sin( ) cos( ) sin( ) cos( ) sin( )n m

m

m i mz zm i m m i m m i mz

cos( ) sin( )

1m i m

Then, as cos(-x) = cos(x) and sin(-x) = -sin(x) we have that

cos( ) sin( ) cos( ) sin( ) cos( ) sin( ).m i m m i m n i n

Therefore, cos( ) sin( )nz n i n for n being a negative integer.


For n being a rational number:

Let , 0pn qq

where p and q are integers.

Then, we have

cos sin cos( ) sin( ) cos sinq

pp pi p i p iq q

Therefore, cos sinp piq q

is one of the values of

cos sin

pqp pi

q q

.

Therefore, the theorem is proved for all rational values of n.

2.3 Solution to the equations of the form

2.3.1 Definition: The nth root of the complex number x + iy is the solutions of the equation nz x iy .

To solve equations of the form nz x iy , we can use

1. A factorization technique or 2. De Moivre’s theorem, together with the result that the n distinct

nth roots of (cos sin )r i are given by 1 2 2cos sin , 0,1,2,..., 1n k kr i k n

n n

That is,1 2 2cos sin , 0,1,2,..., 1n n k kz x iy z r i k n

n n


2.4 Geometrical Representation of the nth roots of a complex number

Geometrically, the nth roots of a complex number can be represented in

an Argand diagram as the vertices of a regular polygon of n sides,

inscribed in a circle of radius r1/n .

The nth root would be spaced at intervals of 2n

from each other.


3. Solutions to equations of the form , where zn

is a real number Consider the equation 1 0nz . It is possible to solve this equation by any of the methods mentioned in §2.3. Hence, we will start analyzing this equation by finding solutions to this equation where n∈ +; using different methods and plot the solutions to this equation on the Argand plane and the unit circle.

3.1 De Moivre’s Theorem and nth roots of unity Given 1nz , we can rewrite it as 1(cos 0 sin 0)nz i 1(cos(0 2 ) sin(0 2 ))k i k

Using De Moivre’s Theorem,

1 0 2 0 21 cos sinn k kz i

n n

2 2cos sink kz in n

, where k = 0, 1, 2, 3, …, (n – 1)

Thus the nth roots of unity are equally spaced around the unit circle with centre at the origin and forming the vertices of a regular n sided polygon.


We express nz as x + 0i when zn equals to x. This means that [cos 0 sin 0] [cos(0 2 ) sin(0 2 )]n nz x i z x k i k

(2 )nz xcis k

1 2 , 0,1,2,...( 1)n kz x cis k nn

Hence, the radius of the inscribed circle would be 1nx and the nth root

would be spaced at intervals of 2n

from each other.

3.3 Roots of the equation zn – 1 = 0 when n = 1

3.3.1 By De Moivre’s theorem:

When n=1, 1 1 0z

11 11 1z z

1 can be expressed in polar form as cos 0 sin 0 0i cis (0 2 )cis k

(2 )cis k .

111 (2 ) (2 )z cis k cis k , where k = 0.

Therefore, the only root of the equation when n = 1, is 1 where = 0. In an Argand plane it will have a real value of 1, and imaginary value of 0. Hence, the solution is (1, 0). In a unit circle, this will be represented as the point (1, 0) on the circumference of the circle, having a radius of 1 unit and the roots of unity would be place at intervals of 0 from each other.

3.3.2 By Factorization When n=1, 1 1 0z and simply, as it can be seen that the equation is in its simplest form, and hence the root of the equation would this be 1.



3.4.1 By De Moivre’s Theorem

When n=2, 2 1 0z 1

2 21 1z z 1 can be expressed in polar form as cos 0 sin 0 0i cis (0 2 )cis k

(2 )cis k .

12 21 (2 )

2kz cis k cis cis k

, where k = 0, 1.

Therefore, the roots of the equation when n = 2 are

at k = 0, 0 0 1z cis cis where = 0.

at k = 1, 1 1z cis cis where = .

In an Argand plane the coordinates are (1, 0) and (-1, 0). In a unit circle, this will be represented as the point (1, 0) and (-1, 0) on the circumference of the circle, having a radius of 1 unit and the square roots of unity would be spaced at intervals of from each other.

3.4.2 By Factorization When n=2, 2 1 0z . Using the difference of square method, we get

2 21 0 ( 1)( 1) 0z z z

1z , 1z

Hence, the roots of the equation are 1 and -1.



When n=3, 3 1 0z 1


(2 )cis k .


13 21 (2 )

3kz cis k cis

, where k = 0, 1, 2.


at k = 0, 2 0 0 13

z cis cis

where = 0.

at k = 1, 2 1 2 1 33 3 2

iz cis cis

where = 2

3 .

at k = 2, 2 2 4 1 33 3 2

iz cis cis

where = 4

3 .

In an Argand plane the coordinates are (1, 0), 1 3,2 2

and

1 3,2 2

. In a unit circle, this will be represented as the points (1, 0),

1 3,2 2

and 1 3,2 2

on the circumference of the circle, having a

radius of 1 unit and the cube roots of unity would be spaced at intervals

of 23

from each other.

3.5.2 By Factorization When n=3, 3 1 0z . Using lemma mentioned in §3.2,

3 21 ( 1)(1 )z z z z 2( 1)( 1) 0z z z

1z Using the quadratic formula,

1 3 1 32 2

1 3 1 32 2

iz

iz




When n=4, 4 1 0z 1


(2 )cis k .

14 21 (2 )

4 2k kz cis k cis cis

, where k = 0, 1, 2, 3.


at k = 0, 0 0 12

z cis cis

where = 0.

at k = 1, 12 2

z cis cis i

where =

2 .

at k = 2, 2 12

z cis cis

where = .

at k = 3, 3 32 2

z cis cis i

where = 3

2 .

In an Argand plane the coordinates are (1, 0), 0,i 1,0 and 0, i . In a

unit circle, this will be represented as the points (1, 0), 0,i 1,0 and

0, i on the circumference of the circle, having a radius of 1 unit and the

quartic roots of unity would be spaced at intervals of 2

from each other.


4 2 31 ( 1)(1 )z z z z z 3 2( 1)( 1) 0z z z z


1z Factorising the cubic,

3 2( 1) 0z z z 2

2

( 1) 1( 1) 0( 1)( 1) 0z z zz z

1z

Using the quadratic formulae,

0 0 42

0 0 42

z i

z i



When n=5, 5 1 0z 1


(2 )cis k .

15 21 (2 )

5kz cis k cis

, where k = 0, 1, 2, 3, 4.


at k = 0, 2 0 0 15

z cis cis

where = 0.

at k = 1, 2 1 2 5 1 2 5 105 5 4

iz cis cis

where = 2

5 .

at k = 2, 2 2 4 5 1 10 2 55 5 4

iz cis cis

where = 4

5 .

at k = 3, 2 3 6 5 1 10 2 55 5 4

iz cis cis

where = 6

5 .


at k = 4, 2 4 8 5 1 2 5 105 5 4

iz cis cis

where = 8

5 .

In an Argand plane the coordinates are (1, 0), 5 1 2 5 10,4 4

5 1 2 5 10,4 4

,

5 1 10 2 5,4 4

,

5 1 10 2 5,4 4

. In a

unit circle, this will be represented as the points (1, 0), 5 1 2 5 10,4 4

5 1 2 5 10,4 4

, 5 1 10 2 5,4 4

and 5 1 10 2 5,4 4

on the circumference of the circle, having a radius of 1 unit and the quintic

roots of unity would be spaced at intervals of 25

from each other.


5 2 3 41 ( 1)(1 )z z z z z z 4 3 2( 1)( 1) 0z z z z z

1z

Expressing the quartic as a product of two quadratics:

4 3 2 2 21 ( 1)( 1)z z z z z az z bz

4 3 2( ) ( 2) ( ) 1z a b z ab z a b z

Comparing the coefficients,

1...[1]a b and 2 1...[2]ab

From equation [1], 1 ...[3]a b

Substituting [3] in [2],


2

2

(1 ) 2 12 1

1 0

b bb b

b b

Applying the quadratic formula,

1 1 4 1 52 2

b

1 5 1 51 12 2

a b

Therefore the quadratics now become,

4 3 2 2 21 5 1 51 1 12 2

z z z z z z z z

Therefore, it can be seen from the quadratics that the solutions to the quadratics are coincident. Hence, the two pairs of quadratic equations that will lead to the other 4 roots of z5 – 1 are,

2 1 5 1 02

z z

and 2 1 5 1 0

2z z

Using the quadratic formula on the above two equations we find the other four solutions,

21 5 1 5 4

2 2 5 1 10 2 52 4

iz

21 5 1 5 4

2 2 5 1 10 2 52 4

iz


21 5 1 5 4

2 2 5 1 2 5 102 4

iz

21 5 1 5 4

2 2 5 1 2 5 102 4

iz

3.8 Representations of the roots of zn – 1 = 0 for n=3, 4 and 5 We are interested in analyzing the geometry of the polygons formed by the roots of the equation and hence the roots for n=1 and n=2 are not shown in the following series of diagrams.

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4. Geometry of the nth roots of

the equation zn – 1=0 Now, we will analyze the polygons produced in §3.8 and will try to formulate a conjecture. A regular polygon is a polygon in which every side has the same length and every angle is the same. For any natural number n ≥ 3, we can draw a regular polygon with n sides. We can ask various questions about a regular n-gon.1 Let us first find the length of the segments produced by drawing a line segment from a root to all other roots as shown in the following figures.

The length of the line segment is found by using the coordinate geometry distance formula, i.e. let d be the distance between two points in a Cartesian plane, then d is given by,

2 22 1 2 1( ) ( )d x x y y where (x1, y1) and (x2, y2) are the two points.

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Applying the sum to product identities, this can be simplified down as following:

2 2

2 2

2

2 2 2 20 0 0 02 sin sin 2 sin cos

2 2 2 2

2 sin sin 2 sin cos

4 sin

m

k k k kn n n nd

k k k kn n n n

kn

2 2 2

2 2 2

2

sin 4 sin cos

4 sin sin cos

4 sin

2 sin

k k kn n n

k k kn n n

kn

kn

where k≥1 and n≠0.

Table 1 Patterns in nth roots of the equation zn – 1=0

Number of sides of the

polygon

Number of line segments from one

root

Nature of roots

Sum of the line

segments

Product of the line

segments

3 2 1 real, 2 conjugate

3.46 3


4.82 4


6.16 5

Observation 2: When two roots are conjugates, then the x-axis will be a line of symmetry for this set of points; and all other conjugate pairs.

Observation 3: It can be seen from Table 1 that, when n is odd, one of the roots will be real (positive or negative) as the other roots are


conjugates of each other and hence, will produce only one real root. When n is even, all the roots can pair off, and might or might not have a pair of real root.

Observation 4: Also, the number of line segments produced from one root to all other roots in a regular n-gon is given by n – 1.

Conjecture: Therefore, the conjecture that can be formed now based on table 1 is that,

The product of the length of the line segments produced from one root to all other roots is equal to n, i.e.

1 2 3 ... nd d d d n where dn is the line segment connecting the nth root

to z=1

More formally this conjecture can be expressed as:

1

12 sin

n

k

knn

where n≠0 and n

Proof:

Step 1: Let 4i

z cis en

Therefore, k

k kz cis cisn n

and

k

k kz cis cisn n

Now,

cos sin cos sin 2 sink k k k k k kz z i i in n n n n

Multiplying both sides by –i:

2 sin 2 sink k k ki z z i in n

Hence the conjecture now becomes,


1

1

nk k

kn i z z

Step 2: Now we can introduce the lemma 2 11 ( 1)(1 ... )n nx x x x x

This lemma can be proved by induction as follows:

Let P(n) be the proposition that 2 11 ( 1)(1 ... )n nx x x x x for

1n .

P(n) is true for n=1 since 1 0 1 2 1 11 1 (1 1)(1 1 1 ...1 ) 0 i.e.

L.H.S=R.H.S

Assume that P(n) is true for n=k, i.e., 2 11 ( 1)(1 ... )k kx x x x x

Therefore, for n=k+1

1

2 1

2 1

2 3

2 3

1( ) 1(( 1)(1 ... ) 1) 1( 1)(1 ... ) 1

( 1)( ... ) ( 1)( 1)(1 ... )

k

k

k

k

k

k

xx xx x x x xx x x x x xx x x x x xx x x x x

Hence, it is showed the proposition P(k+1) is true as 1 2 31 ( 1)(1 ... )k kx x x x x x

Since P(n) is true for n=1, n=k, and n=k+1; by principal of mathematical induction, P(n) is true for all n≥1.

Step 3:Let 2 11( ) 1 ... , 11

nnxf x x x x x

x

where

2 11 ... nx x x is the product of the factors of the equation xn – 1=0 over its (n – 1) roots, i.e. excluding the factor (x – 1) and multiplying the rest of the (n – 1) factors.


Step 4: 21 2

221ki k i

kn n nkx cis e e zn

where 1 1k n as x

cannot be equal to 1 according to f(x).

Now, equating 1

2 1 2

11 ...

nn k

kx x x x z

Therefore, at x = 1, L.H.S = 1+1+1+…+11-1=n. This is to be noted that although f(x) at x=1 might be argued as discontinuous because of the

function 11

nxx

, it can also be argued as continuous 2 11 ... nx x x

is valid over x=1. Therefore, R.H.S = 1

2

11

nk

kz

.

Hence, 1

2

11

nk

kn z

Step 5: Multiplying 1

1

nk

kiz

to both sides of

12

11

nk

kn z

, we get:

1 1 12

1 1 1(1 )

n n nk k k

k k kn iz iz z

12

11

2

11

11

11

1

( ) (1 )

( )

( )

( )

2 sin

nk k

kn

k k k

kn

k k

kn

k k

kn

k

iz z

iz iz

iz iz

i z z

kn


Simplifying, 1

1 2 3 ( 1)

1...

nk n

kiz iz iz iz iz

( 1) (1 2 3 ... ( 1))

12( 1)

121

( 1)12 2

( 1)12 2

( 1)

( 1) ( )

( 1) ( 1)1

n n

n n

n

n n

n

nnn

nn

i z

i z

z

z

Therefore, 1

12 sin

n

k

knn

.

Note that 1n iz cis e and hence, since, the lemma is true by principle of mathematical induction, it is proved by mathematical

deduction that 1

12 sin

n

k

knn

is true where n≠0 and n . This is

also valid for negative value of n as sin( ) sin( ) .

For verification purpose, we test the conjecture for more values of n which is summarized in the following table.

SyC T

n

6 7 8

Av

n=6


Table 2 Verific

n

Also note therified by th

Ahmed ssion Numbe

cation of the

Number osegments 5 6 7

at, all otherhe above.

er: 006644-00

conjecture

of line pro

10.80.7

r observatio

n=8

09

oduct of line

1.73 2 1. 7 1.56 1. 7 1.41 1.8

ons made in

n=

e segments

73 1 6 95 1.95 1 85 2 1.85

n §4.2 stand

=7

1.56 0.87 5 1.41 0.7

ds true as

30

7 77 8


5. Solutions to equations of the form , where zn

is not a real number Consider the equation 0nz i . It is possible to solve this equation by any of the methods mentioned in §2.3 and since, the factorization method has been used in §3 to verify that De Moivre’s theorem is true, we would only use De Moivre’s theorem in this section to find the roots of the equation 0nz i where n∈ +; the solutions to this equation on the Argand plane.

5.1 De Moivre’s Theorem and nth roots of i Given nz i , we can rewrite it as

(4 1)1 cos sin 1 2 12 2 2 2

n kz i cis k cis

Using De Moivre’s Theorem,

11 (4 1)1

2

nn kz cis

(4 1)2

kz cisn

, where k = 0, 1, 2, 3, …, (n – 1)


Thus the nth roots of i are equally spaced around the unit circle with centre at the origin and forming the vertices of a regular n sided polygon.

We express nz as 0 + yi when zn equals to yi. This means that

(4 1)cos sin 22 2 2 2

n kz y i y cis k y cis

(4 1)( )2

n kz ycis

1 (4 1) , 0,1,2,...( 1)2

n kz y cis k nn

Hence, the radius of the inscribed circle would be 1ny and the nth root

would be spaced at intervals of 2n

from each other.

5.2 Roots of the equation zn – i = 0 when n = 1

5.2.1 By De Moivre’s theorem:

When n=1, 1 0z i 1

1 1z i z i

i can be expressed in polar form as cos sin2 2 2

i cis

(4 1)22 2

kcis k cis

.

11(4 1)1

2 2kz cis cis

, where k = 0.

Therefore, the only root of the equation when n = 1, is i where = 2 .

In an Argand plane it will have a real value of 0, and imaginary value of i.




When n=2, 2 0z i 1

2 2z i z i


i cis

(4 1)22 2

kcis k cis

.

12(4 1) (4 1)1

2 4k kz cis cis

, where k = 0,1.


at k = 0, 2 24 2

iz cis

where =

4 .

at k = 1, 5 2 24 2

iz cis

where = 5

4 .

In an Argand plane the coordinates are 2 2,2 2

and 2 2,2 2

.



When n=3, 3 0z i 1

3 3z i z i


i cis

(4 1)22 2

kcis k cis

.

13(4 1) (4 1)1

2 6k kz cis cis

, where k = 0,1,2



at k = 0, 36 2

iz cis

where =

6 .

at k = 1, 5 36 2

iz cis

where = 5

6 .

at k = 2, 96

z cis i

where = 9

6

In an Argand plane the coordinates are 3 1,2 2

, 3 1,2 2

. and 0, i



When n=4, 4 0z i 1

4 4z i z i


i cis

(4 1)22 2

kcis k cis

.

14(4 1) (4 1)1

2 8k kz cis cis

, where k = 0,1,2,3


at k = 0, 8

z cis

where =

8 .

at k = 1, 58

z cis

where = 5

8 .

at k = 2, 98

z cis

where = 9

8


at k = 3, 138

z cis

where = 13

8 .

In an Argand plane the coordinates are cos ,sin8 8

,

5 5cos ,sin8 8

, 9 9cos ,sin

8 8

and 13 13cos ,sin

8 8



When n=5, 5 0z i 1

5 5z i z i


i cis

(4 1)22 2

kcis k cis

.

15(4 1) (4 1)1

2 10k kz cis cis

, where k = 0,1,2,3,4


at k = 0, 2 5 10 ( 5 1)10 4

iz cis

where =

10 .

at k = 1, 510

z cis i

where = 5

10 .

at k = 2, 9 2 5 10 ( 5 1)10 4

iz cis

where = 9

10

at k = 3, 13 10 2 5 ( 5 1)10 4

iz cis

where = 13

10 .

SyC

at

In

5nWthsh


t k = 4, z

n an Argand

2 5 14

10 2 54

5.7 Reprn=3, 4 aWe are interhe roots of hown here i

Figure

Ahmed ssion Numbe

1710

cis

d plane the

0 5 1,4

,

5 1,4

.

esentatind 5 rested in anthe equati

in the follow

3 1,2 2

e 10 Cube Ro

er: 006644-00

10 2

coordinate

10 24

ons of th

nalyzing thion and henwing series o

ots of the equ

09

5 ( 54

i

es are 2

5 5,4

he roots

e geometrynce the rooof diagrams

0, i

uation zn – i=

1) where

5 10 5,4 4

1 0,i and

s of zn –

y of the polyots for n=1 s.

=0 on an Arga

= 1710 .

14

,

d

i = 0 fo

ygons formand n=2 i

3 1,2 2

and Diagram

36

or

med by is not

SyC


9cos ,8

2 5 104

10 2 54

FigDia

Figure 1

Ahmed ssion Numbe

5cos ,s8

9sin8

0 5 1,4

5 1,4

gure 11 Quartagram

12 Quintic Ro

er: 006644-00

5sin8

0

tic Roots of t

oots of the eq

09

cos

0,i

the equation

uation zn – i

c

13 13,sin8

2

10 24

zn – i = 0 on

= 0 on an Ar

cos ,sin8 8

38

2 5 10 5,4 4

5 5 1,4

.

n an Argand

rgand Diagram

37

8

1

,

m


It can be seen from the above argand diagrams that in the case of zn = i,

the roots of the equation also form regular polygons and since they are

in a unit circle, the conjecture would also hold true for the equation zn

= i.


6. Generalizations of the results for zn = x+iy When 1x iy , 2 2 1x y . If the graph of y versus x is plotted, it

would give us a circle of radius 1 unit. This means that the pairs of x and y values of would always give roots that fall on the circumference of a unit circle. As (1,0) and (0,1) are pairs of values of (x, y) that satisfy the equation 2 2 1x y and since the conjecture holds true for these

pairs, it can be generalized from the analysis of the previous sections

that since, the conjecture 1

12 sin

n

k

knn

where n≠0 and n is true

for the roots lying on the circumference of a unit circle, it should also be true for zn = x+iy.

Proof:

Let n=3, cosx , siny , r= 2 2x y

Therefore, 3 2 2 2 , 0z x y cis k k

132 2 2

3kz x y cis

where 0,1,2,3...,( 1)k n

at,


132 2

132 2

132 2

0,3

21,3

42,3

k z x y cis

k z x y cis

k z x y cis

Therefore, the length of the line segment at k = 1, should be

112 sin 2 sin

3 3d

when r=1.

From the above roots, d1 is calculated as follows:

2 22

32 2 2 2cos( ) cos sin( ) sin3 3 3 3md x y

2 2

232 2

1

2232 2

2 2 2 23 3 3 3 3 3 3 32sin sin 2sin cos

2 2 2 2

2sin sin 2sin3 3

d x y

x y

2

232 2 2 2 2 2

232 2 2 2 2

232 2 2

cos3 3

4sin sin 4sin cos3 3 3 3

4sin sin cos3 3 3

4sin3

x y

x y

x y

132 2 2sin

3x y

SyC

A

se

li§4

W

d

pn

k

pre

u

leroar


Assuming th

egment at k

ne segment4.2.

When x iy

md 2x

roof. This m11

12 sin

nn

kr

roduct of tegular polyg

s verify it a

et us take toots of the re noted do

Ahmed ssion Numbe

hat 0 ,

k=1 is 2 sin

t is true, t

1y , 2x

1

2 2 sinn ky

means that

n kn

. S

the line segon formed

as follows:

the equatioequation a

own:

er: 006644-00

and since

3

for n=

he conjectu

2 1y , the

1

2nk rn

, the produ

So does the

gments is by the nth

n 3 3 4z

are plotted

Fi

09

2 2 1x y ,

=3. Since, th

ure must as

length of t

2 sin kn

uct of the l

e conjectur

equal to troots of th

4i . Here, rin an argan

gure 13

, it is show

he general s

s well be t

he segment

as suggeste

ine segment

re still hold

he numberhe equation

2 23 4r

nd plane an

wed that th

statement fo

rue as show

becomes,

ed by the

ts now equ

d true tha

of sides onz x iy

25 5

nd the diag

41

he line

or the

wn in

above

uals to

at the

of the y ? Let

. The gonals


As shown by figure 13, the product of the diagonals 2.96 2.96 8.7616

which is approximately equal to 1 13 3 25 2 sin( ) 5 2 sin( )

3 3

but

not equal to n. Therefore, for our conjecture to be valid r must be 1 and hence the conjecture can be revised as follows:

11

12 sin

nn

k

kn rn

where n≠0 andn and 1r


Conclusion During this portfolio about the nth root of complex numbers, I have discovered how vital it is in real life situations. For instance, in the field of chaos theory, it is used to describe the abrupt movement of waves and particles. In fractal geometry, it can be used to delve into the Mendlebrot Set.

I have used Autograph and Geo Gebra as my graphing software and Microsoft Word for word processing. All in all, this portfolio has explicated and well defined the nature and geometry of the roots of an equation.

JestemTousif

123

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