Download - A GEOMETRIC MODEL OF THE FIELD OF COMPLEX NUMBERS · 9/8/2012 · OF COMPLEX NUMBERS CONSTANTIN LUPU Abstract. In this paper we will construct a geometric model of the –eld of

INTERNATIONAL JOURNAL OF GEOMETRYVol. 1 (2012), No. 2, 65 - 78

A GEOMETRIC MODEL OF THE FIELD

OF COMPLEX NUMBERS

CONSTANTIN LUPU

Abstract. In this paper we will construct a geometric model of the�eld of complex numbers by using the elementary plane Euclidean geometrynotions.

1. Introduction

Considering the set of complex numbers, introduced by the axiomaticmethod, it can be build a geometric model of this set of numbers by usingelementary notions of the plane Euclidean geometry (segment, segment�smeasure, angle, angle�s measure) and also the properties of some elementarygeometric transformations (homothety, rotation, symmetry).

De�nition 1.1 ([2], [3]). The triplet (K;�;�) ; where K 6= ?, is called the�eld of complex numbers if the following conditions (axioms) are true:

I: (K;�;�) is a commutative field;II: The field (K;�;�) is an extension of the �eld of real numbers (R;+; �);III. There exists an element i in K, with these properties:1. i� i = i2 = �1 2 R;2. for every element z in K there exist the real numbers x and y so that:

z = x� i� y.

� � � � � � � � � � � � �Keywords and phrases: complex number, set, model.(2010)Mathematics Subject Classi�cation: 51M04, 51M25.Received: 21.06.2012. In revised form: 4.07.2012. Accepted: 13.08.2012.

66 Constantin Lupu

Remark 1.1. The �eld�s (K;�;�) property of being real numbers �eldextension has the following meaning: the �eld (K;�;�) contains an sub-�eld (I;�;�) which is isomorph with the �eld (R;+; �) through a function' : I ! R, so that '(u) = x, for every u 2 I; the function�s properties allowthe �elds (I;�;�) and (R;+; �) to identify: if u, u1 and u2 are elements inI so that '(u) = x, '(u1) = x2 and '(u2) = x2, then u = x,

u1 � u2 = x1 + x2and

u1 � u2 = u1 � u2;furthermore, each element u 2 I can be replaced by the real number x = '(u)and in the calculations we consider

x� y (in I) = x+ y (in R)and

x� y (in I) = x � y (in R):

Remark 1.2. In the equalities from the �rst de�nition we have

i� i = i2 = �1 2 Rand z = x� i�y; where the real numbers �1; x and y represent, in fact, theuniquely determined elements � = i2, � and from the sub�eld (I;�;�) sothat '(�) = �1 ('(�) = '(i2) = i2 = �1); '(�) = � = x and '( ) = = y.

Constructing a model of complex numbers set means to specify a construc-tion process of a nonempty K set and to endow it with two composition laws(noted, for example, with the symbols � and �) so that the triplet (K;�;�)must verify the conditions I, II and III from De�nition 1.1. In the case of realnumbers set, there are known few processes from which we get the modelsof complex numbers set:1. matrix process (using subsets of the second order quadratic matrix,

with real numbers elements; see [1] or [4]);2. the quadratic expansion process or the process with ordered pairs of

real numbers (based on the quadratic expansion of the �eld of real numbers(R;+; �); see [2] or [4]);3. the factorisation process (based on the ring of polynomial factorisation

with the R[X] with the real coe¢ cients through its prime ideal; see [5]);

We will build, in the following lines, a geometric model of complex num-bers set by using the Euclidean plan�s elementary geometry notions.In the Euclidean plane "2, which is endowed with a Cartesian system of

coordinates (XOY ), we consider the set K = fz j z = ��!OM; M 2 "2g. In

this set we are consider:- the element (the null vector)

��!OO from K is noted with 0K ;

- for each element z =��!OM in K n f0Kg is associated with the unique

real number t 2 0; 2�), where t = arg z = ��\OX;OM

�, where the angle�

\OX;OM�is positively oriented (see Figure 1);

A geometric model of the �eld of complex numbers 67

Figure 1- the number t = arg z is called the reduced argument and the set Argz =

farg z + 2k�; k 2 Zg is called the extended argument for the element z;- the element 0K =

��!OO has the reduced argument undetermined, which

means arg 0K = t, for all t 2 [0; 2�);

The notion of argument allows the next simple remarks for the elementsz1 =

��!OM and z2 =

��!ON in K n f0Kg:

a) z1 and z2 have the same direction and the same way if and only ifarg z1 = arg z2 (see Figure 2);

Figure 2 Figure 3

b) z1 and z2 have opposite directions if and only if arg z2 � arg z1 =� (mod 2�) (see Figure 3);

c) z1 = z2 if and only if jz1j = jz2j and arg z1 = arg z2;

Also, if z =��!OM is an element from Kn f0Kg ; then:

d) M 2 OX+ , arg z = 0 and M 2 OX� , arg z = �;

e) M 2 OY+ , arg z = �2 and M 2 OY� , arg z = 3�

2 .

We will remind the de�nitions of some elementary geometric transforma-tions that will be used in the construction of the proposed model:

68 Constantin Lupu

Let be the Euclidean plane "2, which is endowed with a Cartesian systemof coordinates (XOY ) :

De�nition 1.2. A homothety of center O and ratio (power) k 2 R�+ is thegeometric transformation � : "2 ! "2 which satis�es the following condi-tions:

1) �(O) = O; were O is the origin of the Cartesian system of coordinates(XOY ) ;2) if M 6= O is a point from "2 and M 0 = � (M), then:a) the point M 0 is situated on the ray (OM ;b) jOM 0j = k � jOM j : (see Figure 4).

Figure 4

De�nition 1.3. The rotation of center O and angle positively oriented �is the geometric transformation < : "2 ! "2 which satis�es the followingconditions:

1) < (O) = O, where O is the origin of the cartesian system of coordinates(XOY ) ;2) if M 6= O is a point from "2 and M 0 = < (M), then:a) �

�\MOM 0

�= �;

b) jOM j = jOM 0j : (see Figure 5).

Figure 5


De�nition 1.4. The symmetry with respect to a line d is the geometrictransformation Sd : "2 ! "2 which satis�es the following conditions:

1) Sd (A) = A; 8A 2 d;2) if M =2 d and M 0 = Sd (M), then:a) MM 0 ? db) jMP j = jPM 0j ; where MM 0 \ d = fPg. (see Figure 6).

Figure 6

2. Main Result

Theorem 2.1. In the Euclidean plane "2 endowed with a Cartesian systemof coordinates (XOY ) ; we consider the set of vectors K =

nzjz = ��!OM; M 2 "2

oand the operations � : K �K ! K, � : K �K ! K where its are de�nedas follows:

1. the operation � represents the usual addition of �xed vectors

(1) from plane (XOY ); with the application point at the origin O;

2. a) if z1 =��!OM and z2 =

��!ON are elements from Kn f0Kg, then

z1 � z2 = z =��!OP; where

jzj =��!OP �� = jz1j � jz2j

and

(2) arg z = (arg z1 + arg z2) (mod 2�)

b) if z1 = 0K and z2 6= 0K ; or z1 6= 0K and z2 = 0K ; orz1 = z2 = 0K , then

(3) z1 � z2 = 0K

70 Constantin Lupu

Theorem 2.2. The triplet (K;�;�) represents a model of the set of thecomplex numbers.

Proof. We will show that the triplet (K;�;�) verify the conditions I,II andIII from De�nition 1.1. First ,we indicate the geometrical procedures for theobtaining of the sum and the product of two elements from Kn f0Kg. Letz1 =

��!OM and z2 =

��!ON are elements of the set Kn f0Kg. Then the element

z = z1 � z2 is obtained by parallelogram rule (see Figure 7),

Figure 7

and the element w = z1 � z2 it is obtained in two steps, as follows:- we consider the homothety � of center O and ratio k = jz1j ; and we

obtain the point N 0 = � (N) so that N 0 2 (ON and jON 0j = jz1j � jON j =jz1j � jz2j;- we consider the rotation < of center O and angle of measure arg z1,

and we obtain the point Q = <(N 0) so that jOQj = jON 0j = jz1j � jz2j and��\N 0OQ

�= arg z = �: The vector w =

��!OQ represents the element z1 � z2

because jwj = jz1j � jz2j and argw = (arg z1+arg z2)(mod 2�) (see Figure 8).

Figure 8

So, if we represent in the plane "2 the elements z1 and z2 from Kn f0Kg,then the element z1 � z2 is obtained by the rule of the parallelogram and


the lement z1� z2 is obtained by composing the rotation < by the center Oand angle � = arg z1 with the homotety by center O and ratio k = jz1j.I. The triplet (K;�;�) is a commutative �eld.(K1) : The composition laws � and � are always de�ned on the set K:

if z1 and z2 are elements from K, then, according to rules (1), (2) and (3),we deduce that z1 � z2 and z1 � z2 are elements from K;(K2) : The composition laws � and � are commutative: if z1; z2 2 K;

then

z1 � z2 = z2 � z1

(property of usual vector addition); considering u = z1� z2 and v = z2� z1,we deduce that

juj = jz1j � jz2j = jz2j � jz1j = jvj

and

arg u = (arg z1 + arg z2)mod 2� = (arg z2 + arg z1) (mod 2� = arg v) ;

so u = v.(K3) : The laws � and � are associative: if z1; z2; z3 2 K; then

(z1 � z2)� z3 = z1 � (z2 � z3)

(the associativity of usual vector addition); considering u = (z1 � z2) � z3and v = z1 � (z2 � z3) and applying the laws (1), (2) and (3) we easilyconclude that u= v;(K4) : 1) The element 0K =

��!OO from K is neutral element with respect

to the law �: If z = ��!OM 2 K; then z � 0K = 0K � z = z;2) The element 1K =

��!OU fromK; with U 2 OX+ and

��!OU �� = 1 is neutralelement with respect to the law �: If z 2 K and z0 = z � 1K ; z00 = 1K � z;then ��z0�� = jzj � j1K j = jzj ; ��z00�� = j1K j � jzj = jzj ;

arg z0 = (arg z + arg 1K)mod 2� = (arg z + 0)mod 2� = arg z

and

arg z00 = (arg 1K + arg z)mod 2� = (0 + arg z)mod 2� = arg z:

It follows that z0 = z00 = z; hence z � 1K = 1K � z = z:(K5) : 1) For every z =

��!OM 2 K there exists an opposite element z =

��!OM 0 2 K, with

��!OM 0�� = ��!OM �� and arg��!OM 0 = (arg z + �)mod 2� (see

Figure 9).

72 Constantin Lupu

Figure 9 Figure 10

If z 2 K and z =��!OM; z =

��!OM 0 with jzj = jzj and

argz = (arg z + �)mod 2�;

then the vectors��!OM and

��!OM 0 are collinear, they have tha same length but

opposite directions, so

z � (z) = (z)� z = 0K :

2) For every z =��!OM 2 Knf0Kg there exists an inverse element z�1 =��!

OM0 2 K, with ��z�1�� = 1

jzj

and

arg z�1 = (2� � arg z)mod 2�

(see Figure 10).If w = z � z�1; then

jwj = jzj ��z�1�� = jzj � 1jzj = 1

and

argw =�arg z + arg z�1

�mod2� = (2�)mod 2� = 0;

we get that w = 1K : Similarly we �nd that z�1 � z = 1K :(K6) The law � is distributive over the law �: Let the vectors z1 =

��!OM1,

z2 =��!OM2 and z =

��!OM be from Knf0Kg: We show that

z � (z1 � z2) = (z � z1)� (z � z2):

Representing in the Euclidean plane "2 the vectors z =��!OM; z1 =

��!OM1,

z2 =��!OM2 and z1 � z2 =

��!OP , it is obvious that the quadrilateral OM1PM2

is a parallelogram (see Figure 11).


Figure 11

a) Considering the homothety � of centre O and power jzj, ifM 01 = �(M1),

M02 = �(M2) and we take the parallelogram OM

01QM

02 (see Figure12) with��OM 0

1

�� = jzj � jz1j,��M 0

1Q�� = jzj � jz2j we deduce that the triangles OM1P

(Figure 11) and OM01Q (Figure 12) are similar.

Figure 12We �nd that

jOP jjOQj =

jz1jjzj � jz1j

i.e.jz1 � z2jjOQj =

1

jzj ;

so jOQj = jzj � jz1 � z2j; in these conditions we obtain that Q = �(P );

74 Constantin Lupu

b) We consider the rotation < with the centre O and angle � = arg z. IfM"1 = <(M

01), Q

0= <(Q) and M"

2 = <(M02) it results that the quadrilateral

OM"1Q

0M"2 is a parallelogram (see Figure 13), with

��!OM"

1 = z � z1,��!OQ

0=

z � (z1 � z2) and��!OM"

2 = z � z2.

Figure 13

Applying the addition rule for the vectors��!OM"

1 and��!OM"

2 , it follows that��!OQ0 = (z � z1)� (z � z2), so

��!OQ0 = z � (z1 � z2) = (z � z1)� (z � z2):

II. The �eld (K;�;�) is an extension of the �eld of real numbers (R;+; �).a) We consider the set I � K, with I = fz 2 K j z = ��!OM , M 2 OXg.

We see that z is an element from I if and only if arg z = 0 or arg z = �(excepting the element 0K which has undertermined argument) We simplyshow that the triplet (I;�;�) is an sub�eld of the �eld K because the nextconditions are satis�ed:1) If z and v are elements from I, then z v is an element from I (the

property results by applying the usual rule of vector addition);2) If z =

��!OM and v =

��!ON 6= 0K are elements from I, then the element

z � v�1 is from I; indeed, since v 2 I; v�1 is an element from I because

arg v�1 = (2� � arg v)mod 2� =�(2� � 0)mod 2� = 0;(2� � �)mod 2� = �

if N 2 OX+if N 2 OX�

Then we have

arg z � v�1 = (arg z + arg v�1)mod 2� = 0 or �;

so z � v�1 2 I.


b) Let ' : (I;�;�) �! (R;+; �) be a function de�ned as follows: ifz =

��!OM is an element from I, then

'(z) = x =

8>><>>:jzj =

��!OM �� ;0;

� jzj = ��!OM �� ;

if M 2 OX+if M = Oif M 2 OX�

The function ' is an isomorphism between the �elds (I;�;�) and (R;+; �)because the next conditions are ful�lled:1) ' is a bijective function (acount of the axiom of construction of a

segment with a precised length and of ');2) '(z1 � z2) = '(z1) + '(z2);3) '(z1 � z2) = '(z1) � '(z2) for any z1 and z2 from I.Let�s assume, without restrict the generality, that z1 =

��!OM and z2 =

��!ON ,

are elements from I with the property that M;N 2 OX+. If jz1j = x1 > 0and jz2j = x2 > 0, then

'(z1 � z2) = '(��!OM ��!ON) = '(��!OP );

with P 2 OX+ and��!OP �� = ��!OM �� + ��!ON �� = x1 + x2 (the rule of addition

of the vectors with the extremities on the axis OX). Therefore,

'(z1 � z2) = '(z1) + '(z2):

The property 2) can be proved in the same manner and in the cases M 2Ox+, N 2 Ox�, or M 2 Ox� and N 2 Ox�; also, in these conditions, if

z1 � z2 =��!OM ��!ON =

��!OQ;

with��!OQ�� = ��!OM �� !ON �� = x1 � x2 and

arg��!OQ = (arg

��!OM + arg

��!ON)mod 2� = 0;

it follows that:

'(z1 � z2) = '(��!OQ) = x1 � x2 = '(z1) � '(z2):

Therefore that any element z =��!OM from I can be identi�ed through the

real number x = '(z) (it�s written simpli�ed z = x) and sums, respectiveproducts of the types z1 � z2 and z1 � z2 from I its can identify with sums,respective products of the types x1 + x2 , respective x1 � x2 from R (wewritten z1 � z2 = x1 + x2 and z1 � z2 = x1 � x2). In conclusion , the �eld(K;�;�) it is an extesion of the �eld of the real numbers (R;+; �), the setR thus considered that a subset of K.III. There is an element i in K with the properties:1: i� i = i2 = �1 2 R;2: For every element z from K, it can be indicated the real numbers x

and y, so that z = x� i� y.We consider the element i from K with i =

��!OV , V 2 OY+ so that

jij =��!OV �� = 1 and, obviously, arg i = �

2 .

76 Constantin Lupu

1: Let be��!OP = i� i; then

��!OP �� = ji� ij = jij � jij = 1 � 1 = 1, andarg

��!OP = (arg i+ arg i)mod 2� =

�

2+�

2= �;

the point P is on the axis OX�. We have

'(i� i) = '(i2) = '(��!OP ) = ��!OP �� = �1;

therefore i2 = �1;2: Let be z =

��!OM 2 Knf0Kg. We will demostrate that the real numbers

x and y are existing, so that z = x� i� y;a) We will assume that the extremity M of the vector z =

��!OM is posi-

tioned in the �rst or the second quadrant (see Figure 14) and let M1 andM2 are the projections of M on the axis OX, respectively OY .

Figure 14

According to the rule of addition of the vectors, it results that:

(4)��!OM =

��!OM1 �

��!OM2

Since, the point M1 is situated on the axis OX, the vector��!OM1 it is iden-

ti�ed through the real number x = ��!OM1

��, as M1 2 OX+ or M1 2 OX�,so

(5)��!OM1 = x:

Now we consider the simmetry Sd, where d represents the �rst bisetrix and

let be M02 = Sd(M2) 2 OX (see Figure14), so

��!OM02

�� = ��!OM2

��. In theeseconditions,

��!OM2 = i�

��!OM

02 because

��!OM2

�� = jij � ��!OM 02

�� = ��!OM 02

��, and(arg i+ arg

��!OM 0

2)mod 2� =�

2+ 0 =

�

2= arg

��!OM2:


Since, the vector��!OM

02, is indeti�ed through the real number y =

��!OM02

�� =��!OM2

��, it results that(6)

��!OM2 = i� y:

Replacing (5) and (6) in (4), we deduce that:

z =��!OM =

��!OM1 �

��!OM2 = x� i� y:

b)We assume now that the extremityM of z =��!OM is positioned in the third

or the fourth quadrant (see Figure15). Let M1and M2 be the projections ofthe point M on the axis OX and OY respectively.

Figure 15

Considering M02 = Sd(M2), we deduce that M

02 2 OX� and

��!OM

02 vector

is identi�ed through the real number y < 0, with y = ��!OM

02

�� = � ��!OM2

��.But

��!OM2 = i�

��!OM

02 because

��!OM2

�� = jij � ��OM 0

2

�� andarg(i�

��!OM

02) = (arg i+ arg

��!OM

02)mod 2� = (

�

2+ �)mod 2� =

3�

2:

Since��!OM

02 = y (through ' isomorphism), it results that

��!OM2 = i � y.

Therefore, replacing in the equality��!OM =

��!OM1 +

��!OM2,

��!OM1 with x and��!

OM2 with i� y, we obtain z =��!OM =

��!OM1 �

��!OM2 = x� i� y: �

Being met the �rst, second, and third conditions from the de�nition ofthe complex numbers set, we deduce that the ensemble (K;�;�) is a modelof the complex numbers set and they contain the follow conditions:- K set is noted with the symbol C;- the operation � and � are noted with the classic symbols "+" and "�";- the triplet (C,+,�) is named the complex numbers set;- the elements of C set are named complex numbers;- the �eld (C , +, �) is named the complex numbers �eld

78 Constantin Lupu

Acknowledgement. The author is grateful to the referees for theiruseful suggestions.

References

[1] Alef0, Algebra - Real Numbers, Computation, Complex Numbers (Romanian), E.D.P.Publisher, Bucuresti, 1974.

[2] Beju, A. and Beju, I., Encyclopedia of mathematics (Romanian), Ed. Stiinti�c¼a siEnciclopedic¼a, Bucuresti, 1983.

[3] Berceanu, C., Olympiad problems (Romanian), Gra�t Publisher, Bac¼au, 1996.[4] Lupu, C., Ciofu, C. and Lupu, L., The �eld of complex numbers (Romanian), Gra�t

Publisher, Bac¼au, 2005.[5] Miron, R. and Brânzei, D., Foundations of arithmetic and geometry (Romanian), Ed.

Academiei, Bucuresti, 1983.[6] Sabac, G., Special mathematics, E.D.P. Publisher, Bucuresti, 1975.

"GRIGORE ANTIPA" NATIONAL COLLEGEHENRI COAND¼A 7, BAC¼AU, ROMANIAE-mail address: [email protected]