Complex Numbers from A to…Z

You can free download Complex Numbers from A to…Z by Titu Andreescu • Dorin Andrica in pdf format.

You can free download Complex  Numbers  from A to…Z by Titu Andreescu • Dorin Andrica in pdf format.

About Complex  Numbers  from A to…Z

Definition of Complex Numbers 

In what follows, we assume that the definition and basic properties of the set of real .
numbers R are known.Let us consider the set R2 = R × R = {(x, y)|x, y ∈ R}. Two
elements (x1, y1) and (x2, y2) of R2 are equal if and only if x1 = x2 and y1 = y2.
The operations of addition and multiplication are defined on the set R2 as follows.

z1 + z2 = (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) ∈ R2 and

z1 · z2 = (x1, y1) · (x2, y2) = (x1x2 − y1y2, x1y2 + x2y1) ∈ R2,

for all z1 = (x1, y1) ∈ R2 and z2 = (x2, y2) ∈ R2.

The element z1 + z2 ∈ R2 is called the sum of z1 and z2, and the element

z1 · z2 ∈ R2 is called the product of z1 and z2 and will often be written simply

z1z2. 

Remark 

(1) If z1 =(x1, 0)∈R2 and z2 =(x2, 0)∈R2, then z1z2 =(x1x2, 0).
(2) If z1 = (0, y1) ∈ R2 and z2 = (0, y2) ∈ R2, then z1z2 = (−y1y2, 0).

Examples.
(1) Let z1 = (−5, 6) and z2 = (1, −2). Then
z1 + z2 = (−5, 6) + (1, −2) = (−4, 4) 

Properties Concerning Addition 

The addition of complex numbers satisfies the following properties:
(a) Commutative law.z1 + z2 = z2 + z1 for all z1, z2 ∈ C.
(b) Associative law.(z1 + z2) + z3 = z1 + (z2 + z3) for all z1, z2, z3 ∈ C.
Indeed, if z1 = (x1, y1) ∈ C, z2 = (x2, y2) ∈ C, z3 = (x3, y3) ∈ C, then
(z1 + z2) + z3 = [(x1, y1) + (x2, y2)] + (x3, y3)
= (x1 + x2, y1 + y2) + (x3, y3) = ((x1 + x2) + x3, (y1 + y2) + y3),
z1 + (z2 + z3) = (x1, y1) + [(x2, y2) + (x3, y3)]
= (x1, y1) + (x2 + x3, y2 + y3) = (x1 + (x2 + x3), y1 + (y2 + y3)).
The claim holds due to the associativity of the addition of real numbers.
(c) Additive identity. There is a unique complex number 0 = (0, 0) such that
z + 0 = 0 + z = z for all z = (x, y) ∈ C.

(d) Additive inverse. For every complex number z = (x, y), there is a
unique −z = (−x, −y) ∈ C such that  z + (−z) = (−z) + z = 0.
The reader can easily prove the claims (a), (c), and (d).
The number z1 −z2 = z1 +(−z2) is called the difference of the numbers z1
and z2. The operation that assigns to the numbers z1 and z2 the number
z1 − z2 is called subtraction and is defined by
z1 − z2 = (x1, y1) − (x2, y2) = (x1 − x2, y1 − y2) ∈ C. 

Properties Concerning Multiplication

The multiplication of complex numbers satisfies the following properties:
(a) Commutative law.z1 · z2 = z2 · z1 for all z1, z2 ∈ C.
(b) Associative law.(z1 · z2) · z3 = z1 · (z2 · z3) for all z1, z2, z3 ∈ C.
(c) Multiplicative identity. There is a unique complex number 1 =(1, 0) ∈ C such that
z · 1 = 1 · z = z for all z ∈ C.
A simple algebraic manipulation is all that is needed to verify these equalities:
z · 1 = (x, y) · (1, 0) = (x · 1 − y · 0, x · 0 + y · 1) = (x, y) = z and
1 · z = (1, 0) · (x, y) = (1 · x − 0 · y, 1 · y + 0 · x) = (x, y) = z.
(d) Multiplicative inverse. For every complex number z = (x, y) ∈ C∗,
there is a unique number z−1 = (x′, y′) ∈ C such that z · z−1 = z−1 · z = 1.
To find z−1 = (x′, y′), observe that (x, y) ±= (0, 0) implies x ±= 0 or y ±= 0,
and consequently, x2 + y2 ±= 0. 

The relation z · z−1 = 1 gives (x, y) · (x′, y′) = (1, 0), or equivalently

 { xx′ − yy′ = 1 ,yx′ + xy′ = 0}
x′ = x∕x2 + y2 and y′ = −y∕x2 + y2 and y′ = −y
z−1 = 1∕z =x∕x2 + y2
y ∕x2 + y2 x ∈ C∗. 

Solving this system with respect to x′ and y′, one obtains .hence the multiplicative inverse of the complex number z = (x, y) ∈ C∗ . By the commutative law, we also have z−1 · z = 1. Two complex numbers z1 = (x1, y1) ∈ C and z = (x, y) ∈ C∗uniquely determine a third number, called their quotient, denoted by zz1.

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