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Complex numbers

Written by Zero

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Reminders and historical approach

- Solving a 1st degree equation --> of the form
`ax + b = 0`

(with a ≠ 0) --> always only one solution:`x = -(a/b)`

- Solving a 2nd degree equation --> of the form
`ax² + bx + c = 0`

--> use`x = (-b ±√(b²-4ac)/(2a)`

- 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) = -∞`

- To find the
**integral roots**: Horner's method (1786 - 1837) - 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:

- 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
```

- 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
```

- 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**

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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**

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New law

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

- Law already known about 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

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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

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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 real2 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)*

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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 coordinatesTo 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⁺)

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Square roots of complex numbers

- 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.