Complex numbers C
Luca Zaffonte
From a set point of view, the set of complex numbers is the set of ordered pairs of real numbers (Cartesian product
of R for himself R2 = R × R ). R2 = {(a, b) : a ∈ R, b ∈ R}
On this set we define the following operations:
• Sum: (a, b) + (c, d) = (a + c, b + d)
• Product: (a, b) · (c, d) = (ac − bd, ad + bc)
0.1 Algebraic form of complex numbers
Se x, y ∈ R ⇒
• (x, 0) + (y, 0) = (x + y, 0)
• (x, 0) · (y, 0) = (xy − 0, 0 + 0) = (xy, 0)
I identify these complex numbers (with the second null component) with real numbers, hence the operations of C
operate as the operations of R on the numbers of the type (x, 0).
1 The imaginary unity
The complex number i = (0, 1) it is called an imaginary unit.
If a, b ∈ R ⇒
• i · b = (0, 1) · (b, 0) = (0, b)
• a + ib = (a, 0) + (0, b) = (a, b)
Each ordered pair can be written as the sum of a + ib (called algebraic form). So:
C = {a + ib : a, b ∈ R}
Observation: i2 = i · i = (0, 1)(0, 1) = (−1, 0) = −1. In C the equation x2 + 1 = 0 has at least one solution.
Observation: a + ib = c + id ⇐⇒ (a, b) = (c, d) ⇐⇒ a = c e b = d
Then the algebraic representation of complex numbers is unique. Just because it is unique, I can name a as the
real part ofz and b imaginary part. If the real part of a complex number z equals 0, then z is purely imaginary (or
pure imaginary).
Theorem: (C, +, ·) it is a field.
C is not an ordered field because the equation x2 + 1 = 0 has a solution. in an ordered field it cannot have it.
Observation: C is a field but it is not possible to introduce an order relation that makes it an ordered field.
2 Algebraic calculus in C
• Sum: (a + ib) + (c + id) = (a + c) + i(bd)
• Product: (a + ib) · (c + id) = ac + aid + cib + |{z}
i2 bd = (ac − bd) + i(ad + cb)
=−1
• Reverse: if z = a + ib ∈ C, z ̸= 0((a, b) ̸= (0, 0))
1
, then
1 a − ib a − ib a b
= = 2 2
= 2 +i
a + ib (a + ib)(a − ib) a +b a + b2 a + b2
2
• Quotient:
a + ib (a + ib)(c − id) (ac + bd) + i(bc − ad)
= = =
c + id (c + id)(c − id) c2 + d2
ac + bd bc − ad
= 2 +i 2
c + d2 c + d2
2.1 Graphical representation of complex numbers
Complex numbers can be put in one-to-one correspondence with the points of a Cartesian plane (Argand-Gauss
plane).
2.2 Geometric interpretation of operations in C
1. Given a + ib e c + id ∈ C, their sum is on the vertex of the parallelogram constructed on the sides identified
by the two complex numbers.
2. Multiplication by i, given x + iy ∈ C i · (x + iy) = −y + ix, which is equivalent to a rotation of π
2
2.3 Conjugate complex
Given z = x + iy the complex number z̄ = x − iy is called conjugate of z
3 Properties of the conjugate complex
• It is symmetrical with respect to the real axis
• z1 + z2 = z1 + z2 , ∀z1 , z2 ∈ C
• z1 · z2 = z1 · z2 , ∀z1 , z2 ∈ C
Demonstration: z1 = x1 + iy1 e z2 = x2 + iy2
z1 · z2 = (x1 x2 − y1 y2 ) + i (x1 y2 + x2 y1 )
= (x1 x2 − y1 y2 ) + i (x1 y2 + x2 y1 )
z1 · z2 = (x1 − iy1 ) (x2 − iy2 )
= (x1 x2 − y1 y2 ) + i (−x1 y2 − x2 y1 )
= (x1 x2 − y1 y2 ) − i (x1 y2 x2 y1 ) = z1 z2
• z1
z2 = z1
z2 ∀z1 , z2 ∈ C, z2 ̸= 0
• z + z̄ = 2 Re z∀z ∈ C
• z − z̄ = 2 Im z∀z ∈ C
• z · z̄ = (Rez)2 + (Imz)2 ∀z ∈ C
• z̄ = z∀z ∈ C
• z̄ = z ⇐⇒ Im z = 0 ⇒ z ∈ R
2
Luca Zaffonte
From a set point of view, the set of complex numbers is the set of ordered pairs of real numbers (Cartesian product
of R for himself R2 = R × R ). R2 = {(a, b) : a ∈ R, b ∈ R}
On this set we define the following operations:
• Sum: (a, b) + (c, d) = (a + c, b + d)
• Product: (a, b) · (c, d) = (ac − bd, ad + bc)
0.1 Algebraic form of complex numbers
Se x, y ∈ R ⇒
• (x, 0) + (y, 0) = (x + y, 0)
• (x, 0) · (y, 0) = (xy − 0, 0 + 0) = (xy, 0)
I identify these complex numbers (with the second null component) with real numbers, hence the operations of C
operate as the operations of R on the numbers of the type (x, 0).
1 The imaginary unity
The complex number i = (0, 1) it is called an imaginary unit.
If a, b ∈ R ⇒
• i · b = (0, 1) · (b, 0) = (0, b)
• a + ib = (a, 0) + (0, b) = (a, b)
Each ordered pair can be written as the sum of a + ib (called algebraic form). So:
C = {a + ib : a, b ∈ R}
Observation: i2 = i · i = (0, 1)(0, 1) = (−1, 0) = −1. In C the equation x2 + 1 = 0 has at least one solution.
Observation: a + ib = c + id ⇐⇒ (a, b) = (c, d) ⇐⇒ a = c e b = d
Then the algebraic representation of complex numbers is unique. Just because it is unique, I can name a as the
real part ofz and b imaginary part. If the real part of a complex number z equals 0, then z is purely imaginary (or
pure imaginary).
Theorem: (C, +, ·) it is a field.
C is not an ordered field because the equation x2 + 1 = 0 has a solution. in an ordered field it cannot have it.
Observation: C is a field but it is not possible to introduce an order relation that makes it an ordered field.
2 Algebraic calculus in C
• Sum: (a + ib) + (c + id) = (a + c) + i(bd)
• Product: (a + ib) · (c + id) = ac + aid + cib + |{z}
i2 bd = (ac − bd) + i(ad + cb)
=−1
• Reverse: if z = a + ib ∈ C, z ̸= 0((a, b) ̸= (0, 0))
1
, then
1 a − ib a − ib a b
= = 2 2
= 2 +i
a + ib (a + ib)(a − ib) a +b a + b2 a + b2
2
• Quotient:
a + ib (a + ib)(c − id) (ac + bd) + i(bc − ad)
= = =
c + id (c + id)(c − id) c2 + d2
ac + bd bc − ad
= 2 +i 2
c + d2 c + d2
2.1 Graphical representation of complex numbers
Complex numbers can be put in one-to-one correspondence with the points of a Cartesian plane (Argand-Gauss
plane).
2.2 Geometric interpretation of operations in C
1. Given a + ib e c + id ∈ C, their sum is on the vertex of the parallelogram constructed on the sides identified
by the two complex numbers.
2. Multiplication by i, given x + iy ∈ C i · (x + iy) = −y + ix, which is equivalent to a rotation of π
2
2.3 Conjugate complex
Given z = x + iy the complex number z̄ = x − iy is called conjugate of z
3 Properties of the conjugate complex
• It is symmetrical with respect to the real axis
• z1 + z2 = z1 + z2 , ∀z1 , z2 ∈ C
• z1 · z2 = z1 · z2 , ∀z1 , z2 ∈ C
Demonstration: z1 = x1 + iy1 e z2 = x2 + iy2
z1 · z2 = (x1 x2 − y1 y2 ) + i (x1 y2 + x2 y1 )
= (x1 x2 − y1 y2 ) + i (x1 y2 + x2 y1 )
z1 · z2 = (x1 − iy1 ) (x2 − iy2 )
= (x1 x2 − y1 y2 ) + i (−x1 y2 − x2 y1 )
= (x1 x2 − y1 y2 ) − i (x1 y2 x2 y1 ) = z1 z2
• z1
z2 = z1
z2 ∀z1 , z2 ∈ C, z2 ̸= 0
• z + z̄ = 2 Re z∀z ∈ C
• z − z̄ = 2 Im z∀z ∈ C
• z · z̄ = (Rez)2 + (Imz)2 ∀z ∈ C
• z̄ = z∀z ∈ C
• z̄ = z ⇐⇒ Im z = 0 ⇒ z ∈ R
2
