#Complex numbers

Written by Zero

#Reminders and historical approach

1. Solving a 1st degree equation --> of the form ax + b = 0 (with a ≠ 0) --> always only one solution: x = -(a/b)
2. Solving a 2nd degree equation --> of the form ax² + bx + c = 0 --> use x = (-b ±√(b²-4ac)/(2a)
3. Solving the 3rd degree? --> always at least one solution, and at most 3. ----> PoC why we've always at least one solution f(x) = x³ + x² + x + 1 lim(x->-∞) => f(x) = +∞ and lim(x->+∞) => f(x) = -∞
4. To find the integral roots: Horner's method (1786 - 1837)
5. To approximate non-integer roots: graphical, dichotomous methods....
--> Dichotomy
----> Bolzano's theorem
----> On the interval AB we use Bolzano's theorem; we divide the interval AB until equals a change of sign.

Let the equation x³ - 15x - 4 = 0 be used.
-> with Horner: 4 is found as a solution
In the 16th century, Bombelli solves the following equation in the following way:

1. He writes x in the form a - b -> the equation becomes:
(a-b)³ - 15(a-b)-4=0
a³ - 3a²b + 3ab² - b³ - 15(a-b) - 4 = 0
a³ - 3ab(a-b) - b³ - 15(a-b) - 4 = 0
a³ - (a-b)(3ab+15) - b³ - 4 = 0
1. In this equation, a and b are the unknowns. Bombelli then imagines imposing between a and b u, a link that implies the writing, by posing 3ab + 15 = 0 or b = -5/a Thus he kills two birds with one stone: he has only the unknown a left, and the equation is simplified:
a³ - (-5/a)² - 4 = 0
OR
a⁶ - 4a³ + 125 = 0
1. This equation is a bicarbonate equation. By positing a³=t, it becomes: t² - 4t + 125 = 0. But ρ = -484 = 4 * 121.
To continue his reasoning, Bombelli assumes the existence of an imaginary number
√(-1), denoted later i, of square equal to -1.

--> ρ = 4 * 121 * i² ---> t = (4 ±2*11*i)/2 = 2±11*i
4. From the 1st value of t, he finds a = 2 + i because a³ = (2+i)³ which is equivalent to 2³ + 3*2²*i + 3*2*i² + i³ = 8 + 12*i - 6 - i = 2 + 11*i and thus x = a - b = (2 + i) - (-2 + i) = 4!!!
For Bombelli, i² = -1 was a useful fiction: using this imaginary i (which disappears during the calculation), the real solution was found.

• By imitating Bombelli, we can find solutions to the equation x² + 1 = 0.
• In fact, it is like:
x + 3 = 0; has no solution for the one who only knows the naturals (ℕ)
2x + 5 = 0; has no solution for the one who knows only integers (ℤ)
x² - 2 = 0 ; has no solution for the one who knows only the rationals (ℚ)
x² + 1 = 0 ; has no solution for the one who knows only the reals (ℝ)

ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ

#Towards a new ensemble

Goal: to introduce a set larger than ℝ, in which all elements, even negative ones, admit a root of even index and to exploit these elements as tools for algebra, trigonometry, geometry, physics....: the set of complex numbers

#New law

To "make" a complex number, use 2 real numbers --> use ℝ²

• Internal law + : ℝ² * ℝ² -> ℝ² : ((a,b),(c,d)) ~> (a+c, b+d) Properties => (ℝ², +) = commutative group
• External law * : ℝ * ℝ² -> ℝ² : (r,(a,b)) ~> (ra,rb) Properties => (ℝ,ℝ²,+) = vector space of dimension 2
• New law on ℝ²: internal multiplication . : ℝ² * ℝ² -> ℝ²
Properties => (ℝ²{(0,0)}, -) = commutative group

#Construction of ℂ

Preliminary remark

• (r,o) * (a,b) = (ra,rb) = (a,b) * (r,o) // r and (r,o) are 2 elements of very different sets but play a similar role.
• r * (a,b) = (ra,rb) = (a,b) * r` -> "identify" them by an isomorphism
• Let P = {(a, 0) | a ∈ ℝ} ⊂ ℝ² (represented in the Cartesian plane by the X axis)

• f : P -> ℝ : (a,0) ~> a is a bijection that respects the + law because f((a,0)+(b,0)) = f((a,0)) + f((b,0)) and the * law because f((a,0) * (b,0)) = f((a,0)) * f((b,0))

This is an isomorphism => same properties in the 2 fields (ℝ, +, *) and (P, +, *)
=> each pair of type (a,0) will be noted a (a,0)≠a

• What about the other pairs? (a,b) = (a,0) + (0, b) = (a, 0) + (b, 0) * (0, 1)
so i = new notation for (0,1), and (a,b) will be denoted a + bi

N.B. (0,1) * (0,1) = (-1,0), which is noted i² = -1.

• i is called the imaginary unit (it is a square root of -1)
• A complex number = an element of ℝ² which has changed notation.
• The set of these numbers is noted ℂ

#Operations and calculation rules - properties

• Law +: In ℝ²: (a,b) + (c,d) = (a+c, b+d)
==> In ℂ: a + bi + c + di = a + c + (b + d) * i

• Law *: In ℝ²: (a,b) * (c,d) = (ac - bd, ad + bc)
==> In ℂ: (a + bi) * (c + di) = ac - bd + (ad + bc) * i

• (ℂ,+,*) = fields because (ℝ²,+,*) is one.

• ℝ⊂ℂ because ℝ identified with P which is ⊂ℝ²

• Vocabulary: ∀Z ∈ ℂ, Z is uniquely written a + bi. The real a is called the real part of z -> a = R(z)
The real b is called the imaginary part of z -> b = I(z)

if a = 0 and b ≠ 0, z = bi is pure imaginary
if a ≠ 0 and b = 0, z = a is real

• 2 complex numbers are equal <=> they have the same real part and the same imaginary part i.e. a + bi = c + di <=> a = c and b = d

• 2 complex numbers are conjugate <=> they have the same real part and opposite imaginary parts: -> if z = a + bi, its conjugate = a - bi

• ∀Z ∈ ℂ: zˉz ∈ ℝ (note that ˉz = z conjugate)

#Geometric representation of ℂ

• To represent ℝ²: on the Cartesian plane π0
Each vector (with, at its end, a point) is marked by a pair of coordinates

• To any complex number à + bi corresponds 1! pair
--> 1! vector = vector-image of a + bi, 1! point = point-image of a + bi

• To any vector corresponds 1! complex number = its affix
--> If every element of π0 is marked by its affix, we have the Gaussian plane (or Argand's plane)
(a,b) = component of →v
a + bi = affix of →v
→v = image vector of a + bi
M = image point of a + bi

If →v = image vector of z = a + bi; the norm of →v = ||→v|| = √(a² + b²) = the modulus of z
This is denoted |z|; it is also worth √(z * ˉz) (and corresponds to the notion of absolute value if z ∈ ℝ⁺)

#Square roots of complex numbers

1. Existence and form
• Not all real numbers admit square roots
• Does every complex number admit 2 square roots? To be proved! Let z ∈ ℂ --> z = a + bi with a and b real given
z admits a square root <=> ∃ z' ∈ ℂ verifying z'² = z (*); denoted RC(z) = z' = x + yi with x and y searched
(*) ))> (x + yi)² = a + bi
==> x² - y² = a -> x⁴ + y⁴ - 2x²y² = a²
==> 2xy = b -> 4x²y² = b² --> (x² + y²)² = a² + b²
x² + y² = ±√(a² + b²)

In conclusion, any complex number z admits 2 square roots:
If z ∈ ℝ⁺ , the 2 roots are real and opposite;
If z ∈ ℝ-0 , the 2 roots are complex and opposite;
If z ∈ ℂ \ ℝ , the 2 roots are complex and opposite.