# #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.