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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

i 

1

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


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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 alKhwarizmi, 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.


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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 P (x )  an x n  an 1x n 1  an 2x n 2  ...  a1x 1  a0 , an  0

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 x , y  R, i 2  1}

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.

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Candidate Session Number: 006644-009

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 P ( )  0 i.e. R = 0 then P (x )  (x   )Q(x )  0  (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

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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

P(z )  (z  z1 )Pn 1(z )  0 where Pn-1(z) is itself a polynomial of degree n – 1. By applying fundamental theorem of algebra again, the equation Pn(z)

=

1

0

also

has

a

solution,

say

z2,

so

that

P(z )  (z  z1 )(z  z 2 )Pn 2 (z )  0 where Pn-2(z) is a polynomial of degree n – 2. Continuing in this manner, after n applications we have that

P(z )  (z  z1 )(z  z 2 )...(z  z 2 )P0 (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

P (z )  0  P (z )  0 .

P (z )  an z n  an 1z n 1  an 2z n 2  ...  a1z  a0 so that

P (z )  an z n  an 1z n 1  ...  a1z  a0  an z n  an 1 z n 1  ...  a1 z  a0

n

 an z  an 1 z

n 1



 ...  a1 z  a0  P z



And so, as P (z )  0  P z  0 . That is, z is also a solution. 2.2.5 De Moivre’s Theorem

(r (cos   i sin  ))n  r n (cos   i sin  )n  r n (cos(n )  i sin(n ))  r ncis(n )

Let

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Proof: (By Mathematical Induction) Let P(n) be the proposition that (rcis( ))n  r ncis(n ). For n=1, we have that L.H.S  (rcis( ))1  rcis( )  r 1cis(1   )  R.H.S Therefore, P(n) is true for n=1. Assume now that P(n) is true for n = k, that is, (rcis( ))k  r kcis(k ). Then, for n = k + 1, we have (rcis( ))k 1  (rcis( ))k (rcis( ))

 r kcis(k )(rcis( ))  r k 1cis(k )cis( )  r k 1cis(k   )  r k 1cis((k  1) )

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,

z n  z m 

1 1 1 cos(m )  i sin(m )    m cos(m )  i sin(m ) cos(m )  i sin(m ) cos(m )  i sin(m ) z



cos(m )  i sin(m ) 1

Then, as cos(-x) = cos(x) and sin(-x) = -sin(x) we have that cos(m )  i sin(m )  cos(m )  i sin(m )  cos(n )  i sin(n ).

Therefore, z n  cos(n )  i sin(n ) for n being a negative integer.

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For n being a rational number: Let n 

p ,q  0 where p and q are integers. q

Then, we have q

 p  p  p   cos     i sin      cos(p )  i sin(p )   cos   i sin    q   q  p  p  Therefore, cos     i sin    is one of the values of q  q  p

 p   p  q  cos     i sin     .  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 z n  x  iy . To solve equations of the form z n  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 r (cos   i sin  ) are given by 1     2k  r n  cos  n  

    2k    i sin  n  

   , k  0,1,2,..., n  1 

That is, 1     2k      2k  z n  x  iy  z  r n  cos    i sin  n n    

   , k  0,1,2,..., n  1 


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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 .  2  The nth root would be spaced at intervals of   from each other.  n 

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3. Solutions to equations of , where zn

the form

is a real number Consider the equation z n  1  0 . 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 z n  1 , we can rewrite it as z n  1(cos 0  i sin 0)

 1(cos(0  2k  )  i sin(0  2k  )) Using De Moivre’s Theorem,

  0  2k    0  2k   z  1  cos    i sin   n   n  1 n

  2k    2k   z   cos    i sin   n   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.

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We express z n as x + 0i when zn equals to x. This means that z n  x[cos 0  i sin 0]  z n  x[cos(0  2k  )  i sin(0  2k  )]  z n  xcis(2k  ) 1  2k   z  x n cis   n

  , k  0,1,2,...(n  1)  1

Hence, the radius of the inscribed circle would be x n and the nth root  2  would be spaced at intervals of   from each other.  n 

3.3 Roots of the equation zn – 1 = 0 when n = 1 3.3.1 By De Moivre’s theorem: 1

1 1 1 When n=1, z  1  0  z  1  z  1 1 can be expressed in polar form as cos 0  i sin 0  cis 0  cis(0  2k  )

 cis(2k  ) .





 z  1  cis(2k  )

1 1

 cis(2k  ) , 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, z 1  1  0 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.

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3.4 Roots of the equation zn – 1 = 0 when n = 2 3.4.1 By De Moivre’s Theorem When n=2, z  1  0  z  1  z  1 2

2

1 2

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

 cis(2k ) .





 z  1  cis(2k  )

1 2

 2k    cis    cis k   2 

  , where k = 0, 1.

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

  at k = 1, z  cis 1     cis  1 where

at k = 0, z  cis 0    cis 0  1 where = 0.

= .

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, z 2  1  0 . Using the difference of square method, we get

z 2  12  0  (z  1)(z  1)  0  z  1 , z  1

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

3.5 Roots of the equation zn – 1 = 0 when n = 3 3.5.1 By De Moivre’s Theorem 1

When n=3, z 3  1  0  z 3  1  z  13 1 can be expressed in polar form as cos 0  i sin 0  cis 0  cis(0  2k  )

 cis(2k ) .

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

 z  1  cis(2k  )



1 3

 2k    cis   , where k = 0, 1, 2.  3 

Therefore, the roots of the equation when n = 3 are 20   at k = 0, z  cis    cis 0  1 where = 0.  3 

2 1   2  1  i 3 where at k = 1, z  cis    cis   2  3   3  22   4  1  i 3 where at k = 2, z  cis    cis   2  3   3 

=

=

2 . 3 4 . 3

 1 3 In an Argand plane the coordinates are (1, 0),   ,  and  2 2     1 3  ,  . In a unit circle, this will be represented as the points (1, 0),  2  2    1 3  1 3  ,  and   ,   on the circumference of the circle, having a  2 2   2 2     radius of 1 unit and the cube roots of unity would be spaced at intervals  2  of   from each other.  3 

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

z 3  1  (z  1)(1  z  z 2 )  (z  1)(z 2  z  1)  0 z  1

Using the quadratic formula, 1  3 1  i 3  2 2 1  3 1  i 3 z   2 2 z 

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3.6 Roots of the equation zn – 1 = 0 when n = 4 3.6.1 By De Moivre’s Theorem

When n=4, z  1  0  z  1  z  1 4

4

1 4

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

 cis(2k ) .





 z  1  cis(2k  )

1 4

 2k    k   cis    cis   , where k = 0, 1, 2, 3.  4   2 

Therefore, the roots of the equation when n = 4 are 0   at k = 0, z  cis    cis 0  1 where = 0.  2  1    at k = 1, z  cis    cis    i where  2  2

=

 2

.

2  at k = 2, z  cis    cis   1 where  2 

= .

3    3  at k = 3, z  cis    cis    i where  2   2 

=

 

3 . 2

   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

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

  quartic roots of unity would be spaced at intervals of   from each other. 2

3.6.2 By Factorization

When n=4, z 4  1  0 . Using lemma mentioned in §3.2, z 4  1  (z  1)(1  z  z 2  z 3 )  (z  1)(z 3  z 2  z  1)  0

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z  1

Factorising the cubic, (z 3  z 2  z  1)  0  z 2 (z  1)  1(z  1)  0  (z  1)(z 2  1)  0

 z  1 Using the quadratic formulae, 0  0  4 i 2 0  0  4 z   i 2 z 

3.7 Roots of the equation zn – 1 = 0 when n = 5 3.7.1 By De Moivre’s Theorem 1

When n=5, z 5  1  0  z 5  1  z  15 1 can be expressed in polar form as cos 0  i sin 0  cis 0  cis(0  2k  )  cis(2k  ) .



 z  1  cis(2k  )



1 5

 2k    cis   , where k = 0, 1, 2, 3, 4.  5 

Therefore, the roots of the equation when n = 5 are 20   at k = 0, z  cis    cis 0  1 where = 0.  5 

2 1   2  at k = 1, z  cis    cis    5   5 

5  1  i 2 5  10 where 4

22   4   5  1  i 10  2 5 at k = 2, z  cis  where   cis   4  5   5  23    6   5  1  i 10  2 5 where at k = 3, z  cis    cis   4  5   5 

=

2 . 5 =

=

4 . 5 6 . 5

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2 4    8  at k = 4, z  cis    cis    5   5 

5  1  i 2 5  10 where 4

=

8 . 5

 5  1 2 5  10   In an Argand plane the coordinates are (1, 0),  ,  4  4    5 1 2 5  10    5  1 10  2 5    5  1 10  2 5   , , , . In a  4      4 4 4 4 4   ,  ,   5  1 2 5  10   unit circle, this will be represented as the points (1, 0),  ,  4  4    5 1   5  1 10  2 5  2 5  10    5  1 10  2 5    , and  , , ,  4      4 4 4 4 4       on the circumference of the circle, having a radius of 1 unit and the quintic  2  roots of unity would be spaced at intervals of   from each other.  5  3.7.2 By Factorization

When n=5, z 5  1  0 . Using lemma mentioned in §3.2,

z 5  1  (z  1)(1  z  z 2  z 3  z 4 )  (z  1)(z 4  z 3  z 2  z  1)  0 z  1

Expressing the quartic as a product of two quadratics:  z 4  z 3  z 2  z  1  (z 2  az  1)(z 2  bz  1)  z 4  (a  b)z 3  (ab  2)z 2  (a  b)z  1 Comparing the coefficients,

a  b  1...[1] and ab  2  1...[2] From equation [1], a  1  b...[3] Substituting [3] in [2],


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(1  b)b  2  1  b 2  b  2  1  b2  b  1  0 Applying the quadratic formula, b

1 14 1 5  2 2

a  1  b  1 

1 5 1 5  2 2

Therefore the quadratics now become,    1  5  1  5   z4  z3  z2  z  1  z2    z  1  z 2    z  1  2   2            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,   1  5  z2    z  1   0 and  2       

  1  5  z2    z  1  0  2       

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

z 

z 

2

1  5     2   

1  5    4  2   5  1  i 10  2 5    2 4

1  5     2   

1  5    4  2   5  1  i 10  2 5    2 4

2

Syed Tousif Ahmed

17


z 

z 

2

1  5     2   

1  5    4  2     2

1  5     2   

1  5    4  2     2

5  1  i 2 5  10 4

2

5  1  i 2 5  10 4

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.

Sy yed Tousif Ahmed A C Candidate Sesssion Numbeer: 006644-0009

3.8.1 For n=3 n

 1 3 P  ,   2 2   

 1 3 P  ,   2  2  

Figure 1 Cu ube roots of Unity U on an Argand A Diagrram

 1 3 P  ,   2 2   

 1 3 P  ,   2  2  

Figure 2 Cube roots of U Unity on a Un nit Circle

18


3.8.2 For n=4 n

Figure 3 Quartic Roots of Unity on an a Argand Pllane

Figure 4 Quartic Roots of Unity on a Unit Circle

19

Sy yed Tousif Ahmed A

20

C Candidate Sesssion Numbeer: 006644-0009

3.8.3 For n=5 n

    5  1 , 10  2 5    4 4  

   5  1 , 2 5  10   4  4  

    5  1 ,  100  2 5    4 4  

 5 1 2 5  10   ,   4 4  

Figure 5 Qu uintic Roots o of Unity on an n Argand Pla ane  5  1 2 5  10    ,   4 4  

    5  1 , 10  2 5    4 4  

 5 1 100  2 5   ,   4 4  

   5  1 ,  2 5  10   4  4  

Figure 6 Quintic Q Roots of Unity on a Unit Circle e


21

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,

d  (x 2  x1 )2  (y2  y1 )2 where (x1, y1) and (x2, y2) are the two points.

1

http://www.math.rutgers.edu/~erowland/polygons-project.html [accessed November 20th 2012]


22


4 Lengtth of lin 4.1 ne segme ents in the polyg gons W When n=3, a triangle iss produced as shown in n Figure 2. Let d1 and d2 be th he two line segments shown s by th he red liness and drawn n from roott z=1, too other twoo roots; show wed in Figu ure 7.

d1

d2

Fiigure 7

N Now, using the t roots fou und in §3.5.1, the lenggths d1 and d2 are calcu ulated. T Therefore, 2

2

2

2

 2   2  d1   cos(00)  (cos( )) )    sin(00)  (sin( ))   1.73 3   3    4   4  d2   cos(00)  (cos( ))    sin(00)  (sin( ))   1.73 3   3   W When n=4, a square is produced as a shown in n Figure 4. Let d1, d2 and a d3 be the threee line segmeents shown by the red d lines and drawn from m root z= =1, to other three rootts; showed in i Figure 8.


23


d1

d2

d3

F Figure 8

N Now, using the roots found in §3.6.1, § the lengths d1, d2 and d3 are caalculated. Therefore, T 2

2

2

2

2

2

 2   2  d1   cos(00)  (cos( )) )    sin(00)  (sin( ))   1.41 4  4     4  4   d2   cos(00)  (cos( ))    sin(00)  (sin( ))   2 4   4    6   6  d3   cos(00)  (cos( ))    sin(00)  (sin( ))   1.41 4   4  

W When n=5, a pentagon is produced d as shown in Figure 6. 6 Let d1, d2, d3 an nd d4 be the four line segments s sh hown by thee red lines aand drawn from f rooot z=1, to other four roots; show wed in Figurre 9.


24


d1 d2 d3 d4

F Figure 9

N Now, using the roots found f in §33.7.1, the leengths d1, d2, d3 and d4 are caalculated. Therefore, T 2

2

2

2

2

2

2

2

 2   2  d1   cos(00)  (cos( )) )    sin(00)  (sin( ))   1.18 5  5     4  4   d2   cos(00)  (cos( ))    sin(00)  (sin( ))   1.90 5   5    6   6  d3   cos(00)  (cos( ))    sin(00)  (sin( ))   1.90 5   5    8   8  d4   cos(00)  (cos( ))    sin(00)  (sin( ))   1.18 5   5  

4 Observationss 4.2 O Observatio on 1: The general form mula for the line segmen nt dn, drawn n frrom the rooot z=1 can be b expressed d as followin ng:   2k  dm   cos(00)  cos   n 

2

 2k    n(0)  sin      sin  n  

   

2

Syed Tousif Ahmed

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Applying the sum to product identities, this can be simplified down as following: 2

  2k   0  n dm   2 sin  2      

  2k   0  n  sin  2      

  k   2 sin   n 

   k     2 sin     n

  k  sin    n

2

   2k    0  n     2 sin  2            k  cos    n

   

  2k   0  n  cos  2      

     

2

2

 k  2  k  2  k  2  k   4 sin2   sin    4 sin   cos    n   n   n   n   k  4 sin2   n

  2  k   sin    n

 2  k   cos    n

   

 k   4 sin2    n   k   2 sin    n  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

3

2

4

3

5

4

Nature of roots 1 real, 2 conjugate 2 real, 2 conjugate 1 real, 4 conjugate

Sum of the line segments

Product of the line segments

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

Syed Tousif Ahmed

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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. d1  d2  d3  ...dn  n where dn is the line segment connecting the nth root

to z=1 More formally this conjecture can be expressed as: n 1  k  n   2 sin   k 1  n 

where n≠0 and n  

Proof: i   Step 1: Let z  cis    e 4 n 

k

         k  k Therefore, z   cis     cis   and z   cis     n    n    n  k

k

 k    cis    n 

Now,  k z k  z k  cos   n

  k   i sin    n

  k    cos    n

  k    k    i sin    2i sin     n   n 

Multiplying both sides by –i:   k  i z k  z k  i  2i sin   n 





  k     2 sin     n 

Hence the conjecture now becomes,

Syed Tousif Ahmed Candidate Session Number: 006644-009 n 1



n   i z k  z k k 1

27



Step 2: Now we can introduce the lemma x n  1  (x  1)(1  x  x 2  ...  x n 1 ) This lemma can be proved by induction as follows: Let P(n) be the proposition that x n  1  (x  1)(1  x  x 2  ...  x n 1 ) for n  1.

P(n) is true for n=1 since 11  1  (1  1)(10  11  12...111 )  0 i.e. L.H.S=R.H.S Assume that P(n) is true for n=k, i.e., x k  1  (x  1)(1  x  x 2  ...  x k 1 ) Therefore, for n=k+1 x k 1  1  x (x k )  1  x ((x  1)(1  x  x 2  ...  x k 1 )  1)  1  x (x  1)(1  x  x 2  ...  x k 1 )  x  1  (x  1)(x  x 2  x 3  ...  x k )  (x  1)  (x  1)(1  x  x 2  x 3  ...  x k ) Hence, it is showed the proposition P(k+1) is true as

x k 1  1  (x  1)(1  x  x 2  x 3  ...  x k ) 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 f (x ) 

xn  1  1  x  x 2  ...  x n 1 , x  1 where x 1

1  x  x 2  ...  x n 1 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.

Syed Tousif Ahmed

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i 2k   in  2k   n e Step 4: x  1  cis     e   n   1 n

  

2k

 z 2k where 1  k  n  1 as x

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



Now, equating 1  x  x 2  ...  x n 1   x  z 2k k 1



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

xn  1 , it can also be argued as continuous 1  x  x 2  ...  x n 1 x 1 n 1





is valid over x=1. Therefore, R.H.S =  1  z 2k . k 1

n 1



Hence, n   1  z 2k k 1



n 1

Step 5: Multiplying

 iz

k

k 1

n 1

n 1

n 1

k 1

k 1

k 1

n   iz k   iz k   (1  z 2k ) n 1

  (iz k )  (1  z 2k ) k 1 n 1

  (iz k  iz 2k k ) k 1 n 1

  (iz k  iz k ) k 1 n 1

  i(z k  z k ) k 1

 k    2 sin   k 1  n  n 1

n 1





to both sides of n   1  z 2k , we get: k 1

Syed Tousif Ahmed

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Candidate Session Number: 006644-009 n 1

Simplifying,

 iz

k

 iz 1  iz 2  iz 3  ...  iz (n 1)

k 1

 i (n 1)  z (123 ...(n 1)) 

 i (n 1)  z

 n n 1   2 

   



 ( 1)n 1  z n 1 2

 (1)

n 1 2

 (1) 1

 n n 1   2  

 (z n )

 

(n 1) 2



 (1)

 

(n 1) 2

n 1  k  Therefore, n   2 sin  . k 1  n 

Note that z n  cis  e i  1 and hence, since, the lemma is true by principle of mathematical induction, it is proved by mathematical n 1  k deduction that n   2 sin  k 1  n

also

valid

for

negative

  is true where n≠0 and n   . This is 

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.


30

C Candidate Sesssion Numbeer: 006644-0009 T Table 2 Verificcation of the conjecture

n 6 7 8

Number of o line segments 5 6 7

prooduct of linee segments 1  1.73  2  1.73  1  6 0.87  1.56  1.95  1.95  1.56 1  0.87  7 0.77  1.41  1.885  2  1.855  1.41  0.777  8

A note that, all otherr observatioons made in Also n §4.2 stand ds true as verified by th he above. n=6

n= =7

n=8

Syed Tousif Ahmed

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5. Solutions to equations of , where zn

the form

is not a real number Consider the equation z n  i  0 . 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 z n  i  0 where n∈

+

; the solutions to this equation on

the Argand plane.

5.1 De Moivre’s Theorem and nth roots of i Given z n  i , we can rewrite it as          (4k  1)  z n  1   cos    i sin     1  cis   2k    1  cis   2 2  2  2    

Using De Moivre’s Theorem,

  (4k  1)  z  1   cis  2   1 n

 (4k  1)  z  cis  2n 

1

 n   

  , where k = 0, 1, 2, 3, …, (n – 1) 

Syed Tousif Ahmed

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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 z n as 0 + yi when zn equals to yi. This means that          (4k  1)  z n  y  cos    i sin     y  cis   2k    y  cis   2 2  2  2     (4k  1) )  z n  ycis( 2 1  (4k  1)   z  y n cis   , k  0,1,2,...(n  1) 2n   1 n

Hence, the radius of the inscribed circle would be y and the nth root  2  would be spaced at intervals of   from each other.  n 

5.2 Roots of the equation zn – i = 0 when n = 1 5.2.1 By De Moivre’s theorem: 1

When n=1, z 1  i  0  z 1  i  z  i 1       i can be expressed in polar form as cos    i sin    cis   2 2 2    (4k  1)  cis   2k    cis  2 2  

  (4k  1)  z   1  cis  2  

 . 

1

 1      cis   , where k = 0.  2

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.

Syed Tousif Ahmed

33


5.3 Roots of the equation zn – i = 0 when n = 2 5.3.1 By De Moivre’s Theorem

When n=2, z  i  0  z  i  z  i 2

2

1 2

      i can be expressed in polar form as cos    i sin    cis   2 2 2    (4k  1)   cis   2k    cis  . 2 2   

  (4k  1)  z   1  cis  2  

1

 2  (4k  1)     cis   , where k = 0,1. 4   

Therefore, the roots of the equation when n = 2 are   at k = 0, z  cis    4

2 i 2  where = . 4 2

 5   2  i 2 where at k = 1, z  cis    2  4 

=

5 . 4

 2 2  2 2 , , In an Argand plane the coordinates are   and   .  2 2   2  2    

5.4 Roots of the equation zn – i = 0 when n = 3 5.4.1 By De Moivre’s Theorem 1

When n=3, z 3  i  0  z 3  i  z  i 3       i can be expressed in polar form as cos    i sin    cis   2 2 2    (4k  1)  cis   2k    cis  2 2  

  (4k  1)  z   1  cis  2  

1

 . 

 3  (4k  1)     cis   , where k = 0,1,2 6   

Syed Tousif Ahmed

34


Therefore, the roots of the equation when n = 3 are   at k = 0, z  cis    6

3 i  where = . 6 2

 5   3  i where at k = 1, z  cis    2  6   9  at k = 2, z  cis    i where  6 

=

=

5 . 6

9 6

 3 1 , , In an Argand plane the coordinates are   2 2  

 3 1 ,  . and 0, i   2 2  




When n=4, z 4  i  0  z 4  i  z  i 4       i can be expressed in polar form as cos    i sin    cis   2 2 2    (4k  1)  cis   2k    cis  2 2     (4k  1)  z   1  cis  2  

 . 

1

 4  (4k  1)     cis   , where k = 0,1,2,3 8   

Therefore, the roots of the equation when n = 4 are    at k = 0, z  cis   where = . 8 8  5 at k = 1, z  cis   8

  where 

 9  at k = 2, z  cis   where  8 

= =

5 . 8 9 8



Syed Tousif Ahmed

35


 13  13 at k = 3, z  cis  .  where = 8  8 

      In an Argand plane the coordinates are  cos   , sin    , 8  8     5   5  cos   ,sin   8   8 

   , 

  9   9  cos   ,sin   8   8 

   and 

  13   13  cos   , sin   8   8 


When n=5, z 5  i  0  z 5  i  z  i 5       i can be expressed in polar form as cos    i sin    cis   2 2 2    (4k  1)   cis   2k    cis  . 2 2   

  (4k  1)  z   1  cis  2  

1

 5  (4k  1)     cis   , where k = 0,1,2,3,4 10   

Therefore, the roots of the equation when n = 5 are

  at k = 0, z  cis     10 

2 5  10  i( 5  1)  . where = 10 4

 5  at k = 1, z  cis    i where  10 

=

5 . 10

 9   2 5  10  i( 5  1) where at k = 2, z  cis   4  10 

=

9 10

 13   10  2 5  i( 5  1) 13 at k = 3, z  cis  . where =  4 10  10 

   


36


 17  att k = 4, z  cis    10 

10  2 5  i( 5  1) 17 where = . 10 4

 2 5  10 5 1    , In n an Argand d plane the coordinatees are  ,  4 4        2 5  10 , 5  1  ,  4 4   

    10  2 5 ,  5  1  0,i and d   4 4  

 

 10  2 5   5 1  ,   4 4  .

5 Representations of th 5.7 he rootss of zn – i = 0 fo or n n=3, 4 and 5 W are interrested in an We nalyzing the geometry y of the poly ygons form med by th he roots of the equatiion and hen nce the rooots for n=1 and n=2 is i not sh hown here in i the follow wing series of o diagramss.

 3 1 ,    2 2  

 3 1 ,    2 2  

 0, i 

Figure e 10 Cube Roots of the equ uation zn – i= =0 on an Arga and Diagram


37


  5   5 s   cos   , sin 8    8 

           ccos   , sin    8  8  

  9  cos   8 

  9  , sin    8

   

  13   133  cos   , sin   8   8 

   

Fig gure 11 Quarttic Roots of the t equation zn – i = 0 on n an Argand Dia agram

0   0,i

    2 5  100 , 5  1   4  4  

    10  2 5 ,  5  1    4 4  

 2 5  10 5  1  ,  ,  4 4   

   10  2 5 ,  5  1    4 4  .

Figure 12 1 Quintic Ro oots of the equation zn – i = 0 on an Arrgand Diagram m


38

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.

Syed Tousif Ahmed

39


6.

Generalizations

of

the

results for zn = x+iy When x  iy  1 , x 2  y 2  1 . 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 x 2  y 2  1 and since the conjecture holds true for these pairs, it can be generalized from the analysis of the previous sections n 1  k  that since, the conjecture n   2 sin   where n≠0 and n   is true k 1  n 

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

Let n=3, x  cos  , y  sin  , r= x 2  y 2





Therefore, z 3  x 2  y 2 cis   2k  , k  0 z 

at,



x y 2

2

 cis   32k  where k  0,1,2,3...,(n  1) 1 3

Syed Tousif Ahmed

40


 x  y  cis  3     2  k  1, z   x  y  cis    3     4  k  2, z   x  y  cis    3  k  0, z 

2

2

1 3

2

2

1 3

2

2

1 3

Therefore, the length of the line segment at k = 1, should be d1  2 sin

  1  2 sin   when r=1. 3 3

From the above roots, d1 is calculated as follows:



dm 

d1 



x y 2

x 2  y2





2 3

2



2 3

   cos( )  cos    2   3  3 

     2      2 sin  3 3  2      

2

     2      sin( )  sin  3    3

     2    3  sin  3 2      

2

      2     3     2 sin  3 2          

2 2        2 sin      sin       2 sin      cos            3   3     3   3   





x y





x 2  y2

  4 sin   3   sin  3   4 sin   3   cos  3  





x 2  y2

  4 sin   3    sin  3   cos  3   





x 2  y2

  4 sin   3   





2

x 2  y2

2

2 3

2 3

2 3

2 3

2



2

2

 2 sin   3   1 3

2

2

2

2

2



   

2

   

     2    3  cos  3 2      

2

     

     


41

A Assuming th hat   0 , and since x 2  y 2  1 , it is show wed that th he line   seegment at k=1 is 2 sin   for n=3. = Since, th he general statement s foor the 3

line segmentt is true, the conjectu ure must ass well be true as show wn in §44.2.

x  iyy  1 , x 2  y 2  1 , the length of the segment becomes, W When

dm 



x y 2

2



1 n

1  k   k  n 2 sin   as suggesteed by the above   r  2 sin   n   n 

proof. This means m that, the produ uct of the line segmentts now equ uals to

 n1  k n  r  2 sin  k 1   n n 1

 S does thee conjecturre still hold d true thaat the   . So 

product of the t line segments is equal to the number of sides of o the reegular polyggon formed by the nth roots of th he equation z n  x  iyy ? Let us verify it as a follows: t equation z 3  3  4i 4 . Here, r  32  42  25  5 . The leet us take the rooots of the equation are a plotted in an argan nd plane an nd the diaggonals arre noted doown:

Figure 13


42

As shown by figure 13, the product of the diagonals 2.96  2.96  8.7616 1  13    3 2  which is approximately equal to  5  2 sin( )    5  2 sin( )  but 3   3  

not equal to n. Therefore, for our conjecture to be valid r must be 1 and hence the conjecture can be revised as follows: n 1 1  k  n   r n  2 sin   k 1  n 

where n≠0 and n   and r  1


43

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.