IB MATHS HL POTFOLIO TYPE Patters i Complex Numbers A aalytical paper o the roots of a complex umbers ad its geometry i Syed Tousif Ahmed Cadidate Sessio Number: 0066-009 School Code: 0066 Sessio: May 03
Table of Cotets. Itroductio.... Polyomials over the Complex Field... 3 3. Solutios to equatios of the form, where z is a real umber... 9. Geometry of the th roots of the equatio z =0... 5. Solutios to equatios of the form, where z is ot a real umber... 3 6. Geeralizatios of the results for z = x+iy... 39 7. Coclusio... 3
Cadidate Sessio Number: 0066-009. Itroductio Solvig equatios ad fidig aswers to those equatios has always bee ad will be a mathematicia's past time; as it was for al- Khwarizmi, Cardao, Bombelli, Descartes ad may other mathematical greats. Oly a mathematicia ca realize the divie feelig of solvig ad resolvig a equatio ad givig meaig to it. This paper aalyzes the th root of a complex umber ad also relates to the geometry of complex umbers. It idetifies a cojecture relatig to regular polygos ad discusses how the study of complex umbers raises this cojecture. Lastly, the paper describes some applicatio of the roots of a complex umbers i several areas of study.
Cadidate Sessio Number: 0066-009 3. Polyomials over the Complex Field Before we ca move oto the aalysis of the th root of a complex umber, it is essetial to ote dow some otatios used i this paper, some defiitios required to uderstad the mathematical laguage ad some theorems which will be used i explicatig the paper. However, this paper assumes that the reader is familiar with the maipulatio of complex umbers. The agle measure i this paper is i radias.. Defiitios.. Polyomials: A polyomial fuctio, P(x), is a algebraic expressio that takes the form Px ( ) ax a x a x... ax a, a 0 0 Where the coefficiets a, a -, a -,, a, a 0 are real umbers ad the power,, -, -, are ite gers. The degree of a polyomial, degp(x), is the highest power of x i the expressio... Complex Numbers: The set of complex umbers is deoted by C { z : z x iy, where xy, Ri, } The real part of z, deoted by Re(z) is x. That is, Re(z) = x. The imagiary part of z, deoted by Im(z) is y. That is, Im(z) = y.
Cadidate Sessio Number: 0066-009..3 Polyomial over the complex field: Whe the coefficiets, a, a -, a -,, a, a 0, of the polyomial P(x) are complex umbers, the polyomial is a polyomial over the complex field.. Theorems.. Remaider Theorem: For ay polyomial P(x), the remaider whe divided by (x α) is P(α). Proof: The degree of the remaider R(x) must be less tha the degree of the divisor D(x). Therefore if D(x) has degree =, R(x) has degree = 0 ad is therefore costat. if Px () Dx () Qx () Rad Dx () ( x ) the Px () ( x)() Qx R(where R is a costat) Whe x = α, P( ) ( ) Q( x) R P( ) R i.e. the remaider o divisio of P(x) by (x α) is P(α)... Factor Theorem: (x α) is a factor of P(x) if ad oly if P(α) = 0. Proof: By the remaider theorem, Px () ( x)() Qx Rfor all real x. P( ) R But if P( ) 0i.e. R = 0 the Px () ( x)() Qx 0 i.e. (x α) is a factor of P(x). ( x ) Q( x)..3 Fudametal Theorem of Algebra: Every polyomial equatio of the form P(z) = 0, z C, of degree Q + has at least oe complex root. A polyomial P (z), z C, of degree Q +, ca be
Cadidate Sessio Number: 0066-009 5 expressed as the product of liear factors ad hece, produce exactly solutios to the equatio P (z) = 0. Proof: By factor theorem, if P(z) has a solutio z such that P(z ) = 0, the (z z ) is a factor of P(z). Therefore, we have that Pz () ( zz) P () z 0where P -(z) is itself a polyomial of degree. By applyig fudametal theorem of algebra agai, the equatio P - (z) = 0 also has a solutio, say z, so that P() z ( z z )( z z ) P () z 0where P -(z) is a polyomial of degree. Cotiuig i this maer, after applicatios we have that P( z) ( z z )( z z )...( z z ) P( z) where P 0(z) is a costat. Ad so, we 0 fid that there are solutios, amely z, z,, z to the polyomial equatio P(z) = 0... Cojugate Root Theorem: The complex roots of a polyomial equatio with real coefficiets occur i cojugate pairs. Proof: From the fudametal theorem of algebra we have that P(z) = 0. The, it must be the case that Pz () 0 Pz () 0. Let Pz ( ) az a z a z... az a so that 0 Pz ( ) az a z... az a az a z... az a 0 0 0 az a z... az a P z Ad so, as Pz () 0 P z 0. That is, z is also a solutio...5 De Moivre s Theorem ( r(cos isi )) r (cos isi ) r (cos( ) isi( )) rcis ( )
Cadidate Sessio Number: 0066-009 6 Proof: (By Mathematical Iductio) Let P() be the propositio that ( rcis( )) r cis( ). For =, we have that L.H.S ( rcis( )) rcis( ) r cis( ) R.H.S Therefore, P() is true for =. k k Assume ow that P() is true for = k, that is, ( rcis( )) r cis( k). The, for = k +, we have ( ( )) k k rcis ( rcis( )) ( rcis( )) k rcisk ( )( rcis( )) r r k r cis k cis k k ( ) ( ) cis( k ) cis(( k ) ) Therefore, we have that P(k+) is true wheever P(k) is true. Therefore, as P() is true, by the Priciple of Mathematical Iductio, P() is true for =,,3, For beig a egative iteger: Put = -m, where m is a positive iteger. The, z m cos( m ) isi( m ) z m z cos( m ) isi( m) cos( m) isi( m) cos( m) isi( m) cos( m ) isi( m ) The, as cos(-x) = cos(x) ad si(-x) = -si(x) we have that cos( m ) isi( m) cos( m) isi( m) cos( ) isi( ). Therefore, z cos( ) isi( ) for beig a egative iteger.
Cadidate Sessio Number: 0066-009 7 For beig a ratioal umber: p Let, q 0where p ad q are itegers. q The, we have q p p cos isi cos( p) isi( p) cos isi q q Therefore, cos p p i si is oe of the values of q q p p cos i si q q p q. Therefore, the theorem is proved for all ratioal values of..3 Solutio to the equatios of the form.3. Defiitio: The th root of the complex umber x + iy is the solutios of the equatioz x iy. To solve equatios of the formz x iy, we ca use. A factorizatio techique or. De Moivre s theorem, together with the result that the distict That is, th roots of r(cos isi ) are give by k k r cos isi, k 0,,,..., k k z x iy z r cos isi, k 0,,,..., p
Cadidate Sessio Number: 0066-009 8. Geometrical Represetatio of the th roots of a complex umber Geometrically, the th roots of a complex umber ca be represeted i a Argad diagram as the vertices of a regular polygo of sides, iscribed i a circle of radius r /. The th root would be spaced at itervals of from each other.
Cadidate Sessio Number: 0066-009 9 3. Solutios to equatios of the form, where z is a real umber Cosider the equatioz 0. It is possible to solve this equatio by ay of the methods metioed i.3. Hece, we will start aalyzig this equatio by fidig solutios to this equatio where + ; usig differet methods ad plot the solutios to this equatio o the Argad plae ad the uit circle. 3. De Moivre s Theorem ad th roots of uity Give z, we ca rewrite it as z (cos 0 isi 0) (cos(0 k) isi(0 k)) Usig De Moivre s Theorem, 0 0 z cos k isi k k k z cos isi, where k = 0,,, 3,, ( ) Thus the th roots of uity are equally spaced aroud the uit circle with cetre at the origi ad formig the vertices of a regular sided polygo.
Cadidate Sessio Number: 0066-009 0 We express z as x + 0i whe z equals to x. This meas that z x[cos 0 i si 0] z x[cos(0 k) isi(0 k)] z xcis( k) k z x cis, k 0,,,...( ) Hece, the radius of the iscribed circle would be would be spaced at itervals of from each other. x ad the th root 3.3 Roots of the equatio z = 0 whe = 3.3. By De Moivre s theorem: Whe =, z 0 z z ca be expressed i polar form as cos 0 isi 0 cis0 cis (0 k ) cis( k ). z cis( k ) cis( k ), where k = 0. Therefore, the oly root of the equatio whe =, is where = 0. I a Argad plae it will have a real value of, ad imagiary value of 0. Hece, the solutio is (, 0). I a uit circle, this will be represeted as the poit (, 0) o the circumferece of the circle, havig a radius of uit ad the roots of uity would be place at itervals of 0 from each other. 3.3. By Factorizatio Whe =, z 0ad simply, as it ca be see that the equatio is i its simplest form, ad hece the root of the equatio would this be.
Cadidate Sessio Number: 0066-009 3. Roots of the equatio z = 0 whe = 3.. By De Moivre s Theorem Whe =, z 0 z z ca be expressed i polar form as cos 0 isi 0 cis0 cis(0 k ) cis( k). k z cis k cis cis k ( ), where k = 0,. Therefore, the roots of the equatio whe = are at k = 0, at k =, z cis 0 cis0 where = 0. z cis cis where =. I a Argad plae the coordiates are (, 0) ad (-, 0). I a uit circle, this will be represeted as the poit (, 0) ad (-, 0) o the circumferece of the circle, havig a radius of uit ad the square roots of uity would be spaced at itervals of from each other. 3.. By Factorizatio Whe =, z 0. Usig the differece of square method, we get z z z 0 ( )( ) 0 z, z Hece, the roots of the equatio are ad -. 3.5 Roots of the equatio z = 0 whe = 3 3.5. By De Moivre s Theorem Whe =3, 3 3 3 z 0 z z ca be expressed i polar form as cos 0 isi 0 cis0 cis(0 k) cis( k).
Cadidate Sessio Number: 0066-009 k z cis( k) 3 cis 3, where k = 0,,. Therefore, the roots of the equatio whe = 3 are 0 at k = 0, z cis cis0 where = 0. 3 at k =, at k =, i 3 z cis cis 3 3 i 3 z cis cis 3 3 I a Argad plae the coordiates are (, 0), where =. 3 where =. 3 3, ad 3,. I a uit circle, this will be represeted as the poits (, 0), 3, 3 ad, o the circumferece of the circle, havig a radius of uit ad the cube roots of uity would be spaced at itervals of 3 from each other. 3.5. By Factorizatio Whe =3, z 3 0. Usig lemma metioed i 3., 3 z ( z )( z z ) ( z )( z z ) 0 z Usig the quadratic formula, 3 i 3 z 3 i 3 z
Cadidate Sessio Number: 0066-009 3 3.6 Roots of the equatio z = 0 whe = 3.6. By De Moivre s Theorem Whe =, z 0 z z ca be expressed i polar form as cos 0 isi 0 cis0 cis(0 k ) cis( k). k k z cis( k ) cis cis, where k = 0,,, 3. Therefore, the roots of the equatio whe = are 0 at k = 0, z cis cis0 where = 0. at k =, z cis cis i where =. where =. at k =, z cis cis at k = 3, 3 3 z cis cis i where = 3. I a Argad plae the coordiates are (, 0), 0,i, 0 ad0, i. I a uit circle, this will be represeted as the poits (, 0), 0,i, 0 ad 0, i o the circumferece of the circle, havig a radius of uit ad the quartic roots of uity would be spaced at itervals of from each other. 3.6. By Factorizatio Whe =, z 0. Usig lemma metioed i 3., 3 z ( z )( z z z ) 3 ( z )( z z z ) 0
Cadidate Sessio Number: 0066-009 z Factorisig the cubic, 3 ( z z z ) 0 z ( z ) ( z ) 0 ( z )( z ) 0 z Usig the quadratic formulae, 0 0 z i 0 0 z i 3.7 Roots of the equatio z = 0 whe = 5 3.7. By De Moivre s Theorem Whe =5, 5 5 5 z 0 z z ca be expressed i polar form as cos 0 isi 0 cis0 cis(0 k) cis( k ). k z cis( k) 5 cis 5, where k = 0,,, 3,. Therefore, the roots of the equatio whe = 5 are at k = 0, 0 z cis cis0 5 where = 0. at k =, 5 i 5 0 z cis cis 5 5 where =. 5 at k =, at k = 3, 5 i 0 5 z cis cis 5 5 3 6 5 i 0 5 z cis cis 5 5 where =. 5 where = 6. 5
Cadidate Sessio Number: 0066-009 5 at k =, 8 5 i 5 0 z cis cis 5 5 I a Argad plae the coordiates are (, 0), 5 5 0, where = 8. 5 5 5 0, 5 0 5, 5 0 5,. I a,, uit circle, this will be represeted as the poits (, 0), 5 5 0,, 5 5 0, 5 0 5, 5 0 5 ad, o the circumferece of the circle, havig a radius of uit ad the quitic roots of uity would be spaced at itervals of 3.7. By Factorizatio 5 Whe =5, z 5 0. Usig lemma metioed i 3., from each other. 5 3 z ( z )( z z z z ) 3 ( z )( z z z z ) 0 z Expressig the quartic as a product of two quadratics: 3 z z z z z az z bz ( )( ) 3 z abz ab z abz ( ) ( ) ( ) Comparig the coefficiets, a b...[] ad ab...[] From equatio [], a b...[3] Substitutig [3] i [],
Cadidate Sessio Number: 0066-009 6 ( bb ) b b b b 0 Applyig the quadratic formula, b 5 5 5 a b Therefore the quadratics ow become, 3 5 5 z z z z z z z z Therefore, it ca be see from the quadratics that the solutios to the quadratics are coicidet. Hece, the two pairs of quadratic equatios that will lead to the other roots of z 5 are, 5 z z 0ad 5 z z 0 Usig the quadratic formula o the above two equatios we fid the other four solutios, z 5 5 5 i 0 5 5 5 5 i z 0 5
Cadidate Sessio Number: 0066-009 7 z 5 5 5 i 5 0 z 5 5 5 i 5 0 3.8 Represetatios of the roots of z = 0 for =3, ad 5 We are iterested i aalyzig the geometry of the polygos formed by the roots of the equatio ad hece the roots for = ad = are ot show i the followig series of diagrams.
Cadidate Sessio Number: 0066-009 8 3.8. For =3 3 P, 3 P, Figure Cube roots of Uity o a Argad Diagram 3 P, P 3, Figure Cube roots of Uity o a Uit Circle
Cadidate Sessio Number: 0066-009 9 3.8. For = Figure 3 Quartic Roots of Uity o a Argad Plae Figure Quartic Roots of Uity o a Uit Circle
Cadidate Sessio Number: 0066-009 0 3.8.3 For =5 5 0, 5 5 5 0, 5 0 5, 5 5 0, Figure 5 Quitic Roots of Uity o a Argad Plae 5 5 0, 5 0 5, 5 0, 0 5 5 5 0, Figure 6 Quitic Roots of Uity o a Uit Circle
Cadidate Sessio Number: 0066-009. Geometry of the th roots of the equatio z =0 Now, we will aalyze the polygos produced i 3.8 ad will try to formulate a cojecture. A regular polygo is a polygo i which every side has the same legth ad every agle is the same. For ay atural umber 3, we ca draw a regular polygo with sides. We ca ask various questios about a regular -go. Let us first fid the legth of the segmets produced by drawig a lie segmet from a root to all other roots as show i the followig figures. The legth of the lie segmet is foud by usig the coordiate geometry distace formula, i.e. let d be the distace betwee two poits i a Cartesia plae, the d is give by, d ( x x ) ( y y ) where (x, y ) ad (x, y ) are the two poits. http://www.math.rutgers.edu/~erowlad/polygos-project.html [accessed November 0 th 0]
Cadidate Sessio Number: 0066-009. Legth of lie segmets i the polygos Whe =3, a triagle is produced as show i Figure. Let d ad d be the two lie segmets show by the red lies ad draw from root z=, to other twoo roots; showed i Figure 7. d d Figure 7 Now, usig the t roots foud i 3.5., the legths d ad d are calculated. Therefore, d d cos(0) (cos( )) si(0) (si( )).73 3 3 cos(0) (cos( )) si(0) (si( )).73 3 3 Whe =, a square is produced as show i Figure. Let d, d ad d 3 be the threee lie segmets show by the red lies ad draw from root z= =, to other three roots; showed i Figure 8.
Cadidate Sessio Number: 0066-009 3 d d d 3 Now, usig the roots calculated. Therefore, Figure 8 foud i 3.6., the legths d, d ad d 3 are d d d 3 cos(0) (cos( )) si(0) (si( )). cos(0) (cos( )) si(0) (si( )) 6 6 cos(0) (cos( )) si(0) (si( )). Whe =5, a petago is producedd as show i Figure 6. Let d, d, d 3 ad d be the four lie segmets show by the red lies ad draw from root z=, to other four roots; showed i Figure 9.
Cadidate Sessio Number: 0066-009 d d d 3 d Figure 9 Now, usig the roots foud i 3.7., the legths d, d, d 3 ad calculated. Therefore, d are d d d 3 d cos(0) (cos( )) si(0) (si( )).8 5 5 cos(0) (cos( )) si(0) (si( )).90 5 5 6 6 cos(0) (cos( )) si(0) (si( )).90 5 5 8 8 cos(0) (cos( )) si(0) (si( )).8 5 5. Observatios Observatio : The geeral formula for the lie segmet d, draw from the root z= ca be expressed as followig: d m k cos(0) cos k si (0) si
Cadidate Sessio Number: 0066-009 5 Applyig the sum to product idetities, this ca be simplified dow as followig: d m k k k k 0 0 0 0 si si si cos k k k k si si si cos si k k k k si si cos k k k si si cos k si k si where k ad 0. Table Patters i th roots of the equatio z =0 Number of sides of the polygo Number of lie segmets from oe root 3 3 5 Nature of roots real, cojugate real, cojugate real, cojugate Sum of the lie segmets Product of the lie segmets 3.6 3.8 6.6 5 Observatio : Whe two roots are cojugates, the the x-axis will be a lie of symmetry for this set of poits; ad all other cojugate pairs. Observatio 3: It ca be see from Table that, whe is odd, oe of the roots will be real (positive or egative) as the other roots are
Cadidate Sessio Number: 0066-009 6 cojugates of each other ad hece, will produce oly oe real root. Whe is eve, all the roots ca pair off, ad might or might ot have a pair of real root. Observatio : Also, the umber of lie segmets produced from oe root to all other roots i a regular -go is give by. Cojecture: Therefore, the cojecture that ca be formed ow based o table is that, The product of the legth of the lie segmets produced from oe root to all other roots is equal to, i.e. d d d... d 3 where d is the lie segmet coectig the th root to z= More formally this cojecture ca be expressed as: k si k where 0 ad Proof: i Step : Let z cis e Therefore, k z cis cis k k ad k k k z cis cis Now, k k k k k k k z z cos isi cos isi isi Multiplyig both sides by i: k k k k iz z i isi si Hece the cojecture ow becomes,
Cadidate Sessio Number: 0066-009 7 k k i z z k Step : Now we ca itroduce the lemma x ( x )( x x... x ) This lemma ca be proved by iductio as follows: Let P() be the propositio that. x x x x x ( )(... ) for P() is true for = sice 0 ( )(... ) 0 i.e. L.H.S=R.H.S Assume that P() is true for =k, i.e., k k x ( x )( x x... x ) Therefore, for =k+ x k k xx ( ) x x x x x k (( )(... ) ) xx xx x x k ( )(... ) 3 k ( x )( x x x... x ) ( x ) 3 k ( x )( x x x... x ) Hece, it is showed the propositio P(k+) is true as k 3 k x x x x x x ( )(... ) Sice P() is true for =, =k, ad =k+; by pricipal of mathematical iductio, P() is true for all. Step 3:Let x fx x x x x x ( )..., where x x x is the product of the factors of the equatio x... =0 over its ( ) roots, i.e. excludig the factor (x ) ad multiplyig the rest of the ( ) factors.
Cadidate Sessio Number: 0066-009 8 ik i k Step : x cis e e z caot be equal to accordig to f(x). Now, equatig... k x x x x z k k k where k as x Therefore, at x =, L.H.S = +++ + - =. This is to be oted that although f(x) at x= might be argued as discotiuous because of the x fuctio x, it ca also be argued as cotiuous x x... x k is valid over x=. Therefore, R.H.S = z. k Hece, z k k Step 5: Multiplyig k k iz to both sides of z k k, we get: k k k iz iz ( z ) k k k k k k k k k ( iz ) ( z ) k kk ( iz iz ) k k ( iz iz ) k k iz ( z ) k si k
Cadidate Sessio Number: 0066-009 9 Simplifyig, k iz iz iz iz... iz k 3 ( ) i i ( ) (3... ( )) ( ) ( ) z ( ) ( ) ( z ) ( ) ( ) ( ) z z Therefore, k si k. Note that i z cis e ad hece, sice, the lemma is true by priciple of mathematical iductio, it is proved by mathematical k deductio that si is true where 0 ad. This is k also valid for egative value of as si( ) si( ). For verificatio purpose, we test the cojecture for more values of which is summarized i the followig table.
Cadidate Sessio Number: 0066-009 30 Table Verificatio of the cojecture 6 7 8 Number of lie segmets 5 6 7 product of lie segmets.73. 73 6 0.87.56. 95.95.56 0.87 7 0.77..85.85. 0.77 8 Also ote that, all other observatios made i. stads true as verified by the above. =6 = =7 =8
Cadidate Sessio Number: 0066-009 3 5. Solutios to equatios of the form, where z is ot a real umber Cosider the equatioz i 0. It is possible to solve this equatio by ay of the methods metioed i.3 ad sice, the factorizatio method has bee used i 3 to verify that De Moivre s theorem is true, we would oly use De Moivre s theorem i this sectio to fid the roots of the equatio z i 0 where + ; the solutios to this equatio o the Argad plae. 5. De Moivre s Theorem ad th roots of i Give z i, we ca rewrite it as (k ) z cos i si cis k cis Usig De Moivre s Theorem, (k ) z cis (k ) z cis, where k = 0,,, 3,, ( )
Cadidate Sessio Number: 0066-009 3 Thus the th roots of i are equally spaced aroud the uit circle with cetre at the origi ad formig the vertices of a regular sided polygo. We express z as 0 + yi whe z equals to yi. This meas that (k ) z y cos isi y cis k y cis (k ) z ycis( ) (k ) z y cis, k 0,,,...( ) Hece, the radius of the iscribed circle would be would be spaced at itervals of from each other. y ad the th root 5. Roots of the equatio z i = 0 whe = 5.. By De Moivre s theorem: Whe =, z i 0 z i z i i ca be expressed i polar form as cos isi cis (k ) cis k cis. (k ) z cis cis, where k = 0. Therefore, the oly root of the equatio whe =, is i where =. I a Argad plae it will have a real value of 0, ad imagiary value of i.
Cadidate Sessio Number: 0066-009 33 5.3 Roots of the equatio z i = 0 whe = 5.3. By De Moivre s Theorem Whe =, z i 0 z i z i i ca be expressed i polar form as cos isi cis (k ) cis k cis. (k ) (k ) z cis cis, where k = 0,. Therefore, the roots of the equatio whe = are i at k = 0, z cis where =. at k =, 5 i z cis where = 5. I a Argad plae the coordiates are, ad,. 5. Roots of the equatio z i = 0 whe = 3 5.. By De Moivre s Theorem 3 3 Whe =3, z 3 i 0 z i z i i ca be expressed i polar form as cos isi cis (k ) cis k cis. 3 (k ) (k ) z cis cis, where k = 0,, 6
Cadidate Sessio Number: 0066-009 3 Therefore, the roots of the equatio whe = 3 are 3 i at k = 0, z cis where =. 6 6 at k =, at k =, 5 3 i z cis where = 5. 6 6 9 z cis i 6 where = 9 6 I a Argad plae the coordiates are 3,, 3,. ad 0, i 5.5 Roots of the equatio z i = 0 whe = 5.5. By De Moivre s Theorem Whe =, z i 0 z i z i i ca be expressed i polar form as cos isi cis (k ) cis k cis. (k ) (k ) z cis cis 8, where k = 0,,,3 Therefore, the roots of the equatio whe = are at k = 0, at k =, at k =, z z z cis where =. 8 8 5 cis = 8 where 5. 8 9 cis 8 where = 9 8
Cadidate Sessio Number: 0066-009 35 at k = 3, z 3 cis where = 3. 8 8 I a Argad plae the coordiates are cos,si 8 8, 5 5 cos,si 8 8, 9 9 cos,si 8 8 ad 3 3 cos,si 8 8 5.6 Roots of the equatio z i = 0 whe = 5 5.6. By De Moivre s Theorem 5 5 Whe =5, z 5 i 0 z i z i i ca be expressed i polar form as cos isi cis (k ) cis k cis. (k ) (k ) z cis cis 0 5, where k = 0,,,3, Therefore, the roots of the equatio whe = 5 are at k = 0, at k =, 5 0 i( 5 ) z cis 0 5 z cis i 0 where = 5. 0 where =. 0 at k =, z 9 5 0 i( 5 ) cis 0 where = 9 0 at k = 3, z 3 0 5 i( 5 ) cis 0 where = 3. 0
Cadidate Sessio Number: 0066-009 36 7 at k =, z cis 0 I a Argad plae the coordiates are 5 0 5,, 0 5 5,. 0 0 5 i( 5 ) 5.7 Represetatios of the roots of z i = 0 for =3, ad 5 We are iterested i aalyzig the geometry of the polygos formed by the roots of the equatio ad hece the roots for = ad = is ot show here i the followig series of diagrams. 5 5, where = 7. 0 5 0 5,, 0,i ad 3, 3, 0, i Figure 0 Cube Roots of the equatio z i= =0 o a Argad Diagram
Cadidate Sessio Number: 0066-009 37 5 5 cos,si 8 8 cos,si 8 8 9 9 cos, si 8 8 3 3 cos,si 8 8 Figure Quartic Roots of the equatio z i = 0 o a Argad Diagram 0,i 5 0 5, 5 0 5,, 0 5 5, 0 5 5,. Figure Quitic Roots of the equatio z i = 0 o a Argad Diagram
Cadidate Sessio Number: 0066-009 38 It ca be see from the above argad diagrams that i the case of z = i, the roots of the equatio also form regular polygos ad sice they are i a uit circle, the cojecture would also hold true for the equatio z = i.
Cadidate Sessio Number: 0066-009 39 6. Geeralizatios of the results for z = x+iy Whe x iy, x y. If the graph of y versus x is plotted, it would give us a circle of radius uit. This meas that the pairs of x ad y values of would always give roots that fall o the circumferece of a uit circle. As (,0) ad (0,) are pairs of values of (x, y) that satisfy the equatio x y ad sice the cojecture holds true for these pairs, it ca be geeralized from the aalysis of the previous sectios k that sice, the cojecture si k where 0 ad is true for the roots lyig o the circumferece of a uit circle, it should also be true for z = x+iy. Proof: Let =3, x cos, y si, r= x y 3 Therefore, z x y cis k, k 0 3 k z x y cis where k 0,,,3...,( ) 3 at,
Cadidate Sessio Number: 0066-009 0 k z x y 3 cis 0, 3 k z 3 x y cis, 3 k z 3 x y cis, 3 Therefore, the legth of the lie segmet at k =, should be d si si whe r=. 3 3 From the above roots, d is calculated as follows: 3 d x y cos( ) cos si( ) si m 3 3 3 3 3 d 3 3 3 3 3 3 3 3 x y si si si cos 3 x y si si si 3 3 cos 3 3 3 x y si si si cos 3 3 3 3 3 x y si si cos 3 3 3 3 x y si 3 3 x y si 3
Cadidate Sessio Number: 0066-009 Assumig that 0, ad sice x y, it is showed that the lie segmet at k= is si for = =3. Sice, the geeral statemet for the 3 lie segmet is true, the cojecture must as well be true as show i.. Whe x iyy, x y, the legth of the segmet becomes, d m k x y si r si k as suggested by the above proof. This meas that, the product of the lie segmets ow equals to r si k. So does the cojecture still hold true that the k product of the lie segmets is equal to the umber of sides of the regular polygo formed by the th roots of the equatio z x iy? Let us verify it as follows: 3 let us take the equatio z 3 i. Here, r 3 5 5. The roots of the equatio are plotted i a argad plae ad the diagoals are oted dow: Figure 3
Cadidate Sessio Number: 0066-009 As show by figure 3, the product of the diagoals.96.96 8.766 3 3 which is approximately equal to 5 si( ) 5 si( ) but 3 3 ot equal to. Therefore, for our cojecture to be valid r must be ad hece the cojecture ca be revised as follows: k r si k where 0 ad ad r
Cadidate Sessio Number: 0066-009 3 Coclusio Durig this portfolio about the th root of complex umbers, I have discovered how vital it is i real life situatios. For istace, i the field of chaos theory, it is used to describe the abrupt movemet of waves ad particles. I fractal geometry, it ca be used to delve ito the Medlebrot Set. I have used Autograph ad Geo Gebra as my graphig software ad Microsoft Word for word processig. All i all, this portfolio has explicated ad well defied the ature ad geometry of the roots of a equatio.