lunes, 23 de abril de 2018

From Real to Imaginary

Complex Numbers

Introduction

Natural numbers are appropriate to express how many elements are in a given collection of objects. The set of natural numbers is N ={0, 1, 2, ...}
An equation like x + a = b can be solved in N, if ab.
Example: From x + 7 = 12 it is obtained x = 12 - 7 = 5.

An equation of the form x + a = b can be solved without restrictions in the set of integers  which contains zero, positive and negative numbers: Z = {..., -3, -2, -1, 0, +1, +2, +3, ...}.
Example: From x + +9 = -14 it is obtained x = -14 + -9 = -23.

Usually, positive integers +1, +2, +3, ... are identified with the natural numbers 1, 2, 3, ... due to an isomorphism between them. They behave the same way in several instances. The former equation can be written as x + 9 = -14.

In the set Z, an equation of the form x * c = d can be solved  if c is a factor of d.
Example: From  x * 8 = -40  it is obtained x = -40 ÷ 8 = -5.

Rational numbers can be represented in the form a/b, with aintegers and b ≠ 0. Also they can be represented by means of decimals terminating or periodic.
Examples:  3/8= 0.375. -8/3 = -2.666...

In the set Q of rational numbers, an equation of the form x * c = d  can be solved with the only restriction of c ≠ 0.
Example: From x * (3/4) = -8/7 it is obtained x = (-8/7)*(4/3) = -32/21.

An equation like x2 = 2 has not solution in Q , since √2 is not a rational number. It belongs to I, the set of irrational numbers. An irational number can be represented by means of an infinity not periodic decimal.
Example: √2 = 1.41421356...

The set Q I constitutes R, the set of real numbers.

An equation like x2 + 1 = 0 has not solution in R, since for every x in R,  x * x cannot be negative.

If the history continues in the same way as before, it will be necessary to have a numeric set in which an equation like the former has a solution. It has to include a number whose square is -1.

Such a number cannot be real. So, in analogous way with the rational-irrational opposition, that number has been called the imaginary unit and it has been symbolized with i.

In analogous way with the ordinary opposite meaning for the words positive-negative, rational-irrational, the situation  with real-imaginary has a historical background of initial reject from the mathematical community.

The set C

As it will be seen,the set of complex numbers, designated by C, has a subset of it which is isomorphic with R, in the same way as Z has a subset of it, which is isomorphic with N, namely the positive integers. Also, Q has a subset isomorphic with Z, namely the rationals that can be represented with 1 as denominator. Examples: 3/1, -7/1.

To get acquainted with the set R, it is very helpful to represent it  geometrically by means of a straight line whose points correspond to real numbers.

In analogous way, to get acquainted with the set C, it will be very helpful to have a geometric representation of it. The complex numbers are nicely represented as points in a plane.

The cartesian plane, in which every point can be identified by an ordered pair of real numbers, will be the adopted representation for C.

The x-axis of the cartesian representation contains the real component of a complex number. The y-axis contains the imaginary part of a complex number.

A complex number z will be represented as z = a + bi.
The corresponding  point z is determined by the  cartesian coordinates (ab).
Examples: w = 3 + 4i , s = -2 + 3i.

Vectors and complex numbers

Cartesian representation of complex numbers  is connected with representing them by means of vectors in the cartesian plane, since each point (a, b) can be associated with an oriented segment with origin in the point (0, 0) and end-point in (ab).

Examples: w = 3 + 4s = -2 + 3i.


Each oriented segment in the cartesian plane is a representative of a vector in the plane. This means that, for instance, the oriented segment O→W above,  represents a vector in the plane which consists in all of the segments in the plane with same lenght, same direction (parallel) and same orientation as O→W.

In this way, the oriented segment F→G represents the same vector as O→W.

The oriented segments with origin O will be called principal representatives of vectors in the cartesian plane.

In this way, every principal representative of a vector in the cartesian plane can be identified with a complex number.

Addition and subtraction in C

If (a, b) and (c,d) are two points in the cartesian plane then, by definition, 

(a, b) + (c, d) = (a + c, b + d)

Example: (2, 3) + (-3, 1) = (-1, 4)

This can be seen also as addition between the corresponding vectors.

To obtain the result of this addition it can be used the above parallelogram construction, in which the diagonal gives the answer.
Using complex numbers for the above example, (2 + 3i ) + (-3 + 1i) = -1 + 4i.

Subtraction is defined in terms of addition:

                                          (a, b) - (c, d) = (a, b) + (-c, -d


Example: (2 + 3i) - (-3 + 1i)  =  (2 + 3i) + (3 - 1i) = (5 + 2i)

To find a representative for the difference between two vectors, it can be used a graphic rule: "complete the triangle" as it can be seen in above example.

Modulus of a complex number

By definition, for the complex number  z = a + bi the modulus of  z is the real number 

Example: if  z = (3,4) then |z| = 5

It follows that for a given z, all the points in the circle centered in (0, 0) have the same modulus as z.

In other words, the set of  complex numbers z such that |z| = r, for a given positive real number r is a circle centered in (0, 0) and radius r in the cartesian plane.

Multiplication in C

The product between two complex numbers can be performed in the same way as it is done in R, between two algebraic expressions, taking in consideration that i *i = -1. 

If z = a + band w = c + di, then z * w = a.c - b.d + (a.d bc)i

Or, by definition, (a, b) * (c, d) = (a.c - b.da.d + b.c)

Example: (2, 3) * (-3, 1) = (-9, -7)

The imaginary unit i can be identified with the complex number 0 + 1i. So, the correspondent point in the cartesian plane is (0, 1).

Using above definition, (0, 1) * (a, b) = (-b, a). This can be written as i * (a + bi)  = -b + ai.


Example: (0, 1) * (2, -3) = (3, 2)

An important relation between z and can be seen graphically as follows:


By geometric considerations it can be seen that multiplying (a, b) by i determines  a counter-clockwise rotation of 90°, around the point (0, 0), of the vector corresponding to (a, b).

Neutral and inverses in C

If (x, y) is  neutral element for addition in C, then (a, b) + (x, y) = (a, b). 
That is, a + x = a and b + y = b. Conclusion, x = 0 and y = 0. The complex number (0, 0) is neutral for addition.

If z = (a, b)  then its inverse with respect to addition has to be z' = (x, y) such that z + z' = (0, 0). That is, a + x = 0 and b + y = 0. 
Conclusion, x = -a and y = -b.

Example: if z = (-12, 5) then z' = (12, -5), since zz'= (0, 0).

In this case, z' is ordinary designated by -z.

If (x, y) is  neutral element for multiplication in C, then (a, b) * (x, y ) = (a, b). Using above definition, (a, b) * (x, y ) = (ax - by, ay + bx). This means,

ax - by = a
bx + ay = b

The solution for above system is, x = 1, y = 0.

Example: (-8, 9) * (1, 0) = (-8, 9)

If z = (a, b)  then its inverse with respect to multiplication has to be z' = (x, y) such that z * z' = (1, 0). Using definition for multiplication, again, the system to be solved is

ax - by = 1
bx + ay = 0
The solution is

x = a/(a2 +b2 )

y = -b/(a2 +b2 )

Where a+bhas to be different from zero. This means that (0, 0) is the only complex number that has not inverse with respect to multiplication.

Example: if z = (2, 3), then z' = (2/13, -3/13).

In this case, z' is ordinary designated by 1/z.

An equation like Az + B = D, has solution in C, for any complex numbers A, B, D, and A ≠ (0, 0).

Example: (2, -3)z + (-4, 5) = (2/3, -8)
Solution: z = [(2/3, -8) - (-4, 5)] ÷ (2, -3) = (14/3, -13) * (2/13, 3/13) = (145/39, -12/13)

Trigonometric representation


For a complex number z =  (a, b)  for which |z| = r and


the point corresponding to z = (a, b) can be determined by

a = r.cos θ
b = r.sin θ

where θ is the oriented angle with origin in the positive x-axis and end in the line between (0,0) and the point (a, b).

The representation a + bi takes the form r.(cos θ + i.sin θ) which can be simplified as r.cis θ.

Example: 


z = -3 + 4i

tan θ = 4/-3

arctan (4/-3) = 126° 52' 11.63'' = θ

r = 5 

z = 5.cis θ


Multiplication and trigonometric representation

If z = r.cis θ and w = s.cis β then the product z *w can be obtained as follows:

z *w = [r(cos θ + i.sin θ)] * [s(cos β + i.sin β] = 

(r s)[(cos θ * cos β - sin θ * sin β) + i(cos θ * sin β  + sin θ * cos β)] =

(rs)[cos (θ + β) + i sin (θ + β)] = (rs)cis (θ + β)

Example: z = 3.cis 30°. w = 2 cis 45°. z * w = 6 cis 75°


De Moivre's Formula

Multiplication of a complex number by itself  can be extended (by induction) more than twice to obtain the so called De Moivre's Formula.

If z = r cis θ then zn = rn cis nθ

Example: z = 2 cis 30°;  z3 = 8 cis 90°


Roots for the equation zn = w

If the De Moivre's formula were extended for rational exponents then it would be possible to solve an equation like z4 = (0, -1), that is, z4 = 1 cis π.
In this case, w = 1. cis π and z = 11/4 cis (π/4) or simply, cis π/4. 

An equation witth grade 4 has 4 solutions, according with the fundamental theorem of algebra.

To obtain all such solutions is necessary to consider not only the value π/4 but (π/4 + 2n π)/4, for n = 0, 1, 2, 3.

For n = 0, the corresponding solution is z(0) =  cis π/4
For n = 1, the corresponding solution is z(1) =  cis 3π/4
For n = 2, the corresponding solution is z(2) =  cis 5π/4
For n = 3, the corresponding solution is z(3) =  cis 7π/4

These solutions can be represented graphically as follows: