Complex Numbers: Properties, Operations, and Roots, Slides of Mathematics

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CHAPTER 8

THE COMPEX NUMBER

INTRODUCTION:

The system ℂ of complex numbers is the number system of ordinary algebra. It is the smallest set in which, for example, the equation x² = a can be solved when a is any element of ℝ. In our development of the set ℂ, we begin with the product set ℝ ℝ x. The binary relation “=’’ requires (a,b) = (c,d) If and only if a = c and b = d Now each of the resulting equivalence classes contains but a single element. Hence, we shall denote a class (a,b) rather than as [a,b] and so, hereafter, denote ℝ ℝ x by ℂ

8.2 PROPERTIES OF COMPLEX NUMBERS The real numbers are a proper subset of the complex numbers ℂ. For, if in ( ) and ( ⅈ ⅈⅈ) we take b = d = 0, we see that the first components combine exactly as do the real numbers a and c. Thus the mapping a←→ (a,0) is an isomorphism of ℝ onto a certain subset {(a,b):a ϵ ℝ,b = 0} of ℂ. Definition 8.1: The elements (a,b) ϵ ℂin which b ≠ 0, are called imaginary numbers and those imaginary numbers (a,b) in which a = 0 are called pure imaginary numbers. Definition 8.2: For each complex numbers z= (a,b), we define the complex number ͞͞ z = ͞͞͞͞͞͞͞͞͞ (͞͞a͞,͞b) =(a,-b) to be the conjugate of z. Clearly, every real number is its own conjugate while the conjugate of each pure imaginary is its negative. there follow easily Theorem I. The sum (product) of any complex number and its conjugate is a real number. Theorem II. The squar of every pure imaginary number is a negative real number. See also Problem 8.2. The special role of the complex number (1,0) suggests an investigation of another, (0,1). We find (x,y).(0,1) = (-y,x) for every (x,y) ϵ ℂ And in particular, (y,0).(0,1) = (0,1).(y,0) (0,y)

Moreover, (0,1)² = (0,1).(0,1) = (-1,0)←→ -1 in the mapping above so that (0,1) is a solution of z² = -1. Defining (0,1) as the pure imaginary unit and denoting it by i,we have ⅈ² = - And for every (x,y) ϵ ℂ, (x,y) = (x,0) + (0,y) = (x,0) + (y,0).(0,1) = x + yⅈ In this familiar notation, x is called the real part and y is called the imaginary part of the complex number. We summarize: the negative of z = x + y ⅈ is –z = -(x + y ) = -x-yⅈ ⅈ

the conjugate of z = x + y ⅈ is ͞z = ͞͞ x͞͞͞͞+͞y͞ ⅈ = x-yⅈ

for each z = x + y , z.ⅈ ͞z = x² + y² ϵ ℝ

8.3 Subtraction and division on ℂ Subtraction and division on ℂare defined by ( ⅈⅈⅈ) z-w = z + (-w), for all z, w ϵ ℂ ( v) zⅈ ÷ w = z.w⁻¹, for all w ≠ 0, z ϵ ℂ

If Ɵ is the positive angle which O P makes the positive x-axis, we have x = r cos Ɵ, y = r sin Ɵ whence z = x + y ⅈ = r (cos Ɵ + ⅈsin Ɵ) difinition 8.3: The quantity r (cos Ɵ + ⅈsin Ɵ) is called the trigonometric form (polar form) of z. difinition 8.4: the non-negative real number r = |z| = √z.͞z = √x² + y² Is called the modulos (absolute value) of z, and Ɵ is called the anle (amplitude or argument) of z. Now Ɵ satisfies x = r cos Ɵ, y=r sin Ɵ, tan Ɵ=y/x and any two of these determine Ɵ up to an additive multiple of 2π. Usually we shall choose as Ɵ the smallest positive angle.(Note: When p is at o, we have r = 0 and Ɵ arbitrary.) EXAMPLE 1. Express (a) 1+ ⅈ, (b) - √3+ ⅈin trigonometric form. (a) we have r= √1+1= √2 Since tan Ɵ = 1 and cos Ɵ=1/√2, we take Ɵ to be the first quadrant angle 45˚= π/4. thus, 1+ ⅈ= √2(cos π/4+ ⅈsin π/4). (b) here r= √3+1=2, tan Ɵ=-1/√3 and cos Ɵ=-½√3. taking Ɵ to be the second quadrant angle 5π/6, we have

• √3+ ⅈ = 2(cos 5π/6+ ⅈsin 5π/6)

It follows that two complex numbers are equal if and only if their absolute values are equal and their angles differ by an intergral multiple of 2π, i.e.., are congruent modulo 2 π. THEOREM iii. The absolute value of the product of two complex numbers is the product of their absolute values, and the angle of the product is the sum of their angles; and THEOREM iv. The absolute value of the quotient of two complex numbers is the qoutient of their absolute values, and the angle of the quotient is the angle of the numerator minus the angle of the denominator. EXAMPLE 2. (a) When z₁=2(cos ¼π + ⅈsin ¼π ) and z₂=4(cos ¾π + ⅈsin ¾π ) we have z₁. z₂ = 2(cos ¼π + ⅈ sin ¼π ). 4(cos ¾π + ⅈsin ¾π ) =8(cos π + ⅈsin π) = - z₂/z₁ = 4(cos ¾π + ⅈ sin ¾π )÷ 2(cos ¼π + ⅈsin ¼π ) = 2(cos ½π +ⅈ sin ½π) =2ⅈ z₁/z₂ = 2(cos ¼π + ⅈ sin ¼π ) ÷ 4(cos ¾π + ⅈ sin ¾π ) = ½(cos 3π/2 + ⅈ 3π/2) = -½ⅈ

The number of distinct roots are the number of non-cotermal angles of the set {Ø/n + 2kπ/n}. For any positive integer k = nq + m, 0 ͟ < m < n, it is clear that Ø/n + 2kπ/n and Ø/n + 2mπ/n are coterminal. Thus, there are exactly n distinct roots, given by p n[cos(Ø/n + 2kπ/n) +⅟ ⅈ sin(Ø/n + 2kπ/n)], k = 0,1,2,3,......n – 1 These n roots are coordinates of n equispaced poins on the circle, centered at the origin, of radius ⁿ√|Α|. If then z = ⁿ√|Α| (cos θ + ⅈsin θ) is any one of the nth roots of A, the remaining roots may be obtained by successively increasing the angle θ by 2π/n and reducing modulo 2π whenever the angle is greater than 2π. EXAMPLE 3. (a) One root of z⁴ = 1 is r₁ = 1 = cos 0 + ⅈsin 0. increasing the angle successively by 2π/4 = π/2, we find r₂ = cos ½π + ⅈ sin ½π, r₃ = cos π + ⅈ sin π, and r₄ = cos ³/₂π + ⅈsin ³/₂π. Note that had we begun with another root, say -1 = cos π + ⅈ sin π, we would obtain cos ³/₂π +ⅈ sin ³/₂π, cos 2π + ⅈ sin 2π = cos 0 + ⅈ sin 0, and cos ½π + ⅈ sin ½π. These are, of course, the roots obtained above in a different order.

(b) One of the roots of z⁶ = -4√3 -4 ⅈ = 8(cos 7π/6 + ⅈ sin 7π/6) is r₁ = √2(cos 7π/36 + ⅈ sin 7π/ Increasing the angle successively by 2π/6, we have r₂ = √2(cos 19π/36 + ⅈ sin 19π/ r₃ = √2(cos 31π/36 + ⅈ sin 31π/ r₄ = √2(cos 43π/36 + ⅈ sin 43π/ r₅ = √2(cos 55π/36 + ⅈ sin 55π/ r₆ = √2(cos 67π/36 + ⅈ sin 67π/ As aconsequence of Theorem V, we have THEOREM VI. The n nth roots of unity are p = cos 2π/n + ⅈ sin 2π/n p², p³, p⁴,.....pⁿ ¹, pⁿ = 1⁻