Numeri complessi in diverse forme

Potenze

Si ha:

• i⁰ = 1
• i¹ = i
• i² = -1
• i³ = -i
• i⁴ = 1
• i⁵ = i
• i⁶ = -1
• i⁷ = -i

In generale i^m = i^r con r = resto di m/4 e m ∈ Z > 0

m = 4q + r; 0 ≤ r < 4

Esempi

• i¹⁰⁰ = i⁰ = 1
• 100 = 4.25 + 0
• i²²³ = i³ = -i
• 223 = 4.55 + 3
• i¹⁰⁷⁰ = i² = -1
• 1070 = 4.267 + 2

Potenza di un binomio

(a + b)^m = ∑ _k=0^m ( ^m /_k ) a^m-k b^k

( ^m /_k ) = ^m! / (k!(m-k)!)

(x + uy)^m = ∑ _k=0^m ( ^m /_k ) x^m-k (uy)^k, m ∈ Z > 0

:2^-m = 1/2^m m ∈ Z > 0

Esempio

• 2 = 3 + 2i
• (3 + 2i)⁵ = ∑ _k=0⁵ ( ⁵ /_k ) 3^5-k (2i)^k = 0
• m ∈ Z ≥ 0
• ( ⁵ / ₀ ) 3⁵ (2i)⁰ + ( ⁵ / ₁ ) 3⁴ (2i)¹ + ( ⁵ / ₂ ) 3³ (2i)² + ( ⁵ / ₃ ) 3² (2i)³ + ( ⁵ / ₄ ) 3¹ (2i)⁴ + ( ⁵ / ₅ ) 3⁰ (2i)⁵ = 1.243 + 1 + 5.81 2i + 10.27(-1) + 10.5
• (8i) + 5.3 - 16 + 1 - 32i = -15.97 + 12.21i

POTENZE

i₀ = 1i₁ = ii₂ = -1i₃ = -ii₄ = 1i₅ = ii₆ = -1i₇ = -i

In generale i^m = i^r con r = resto di ^m/₄ e m ∈ ℤm = 4.q + r ; 0 ≤ r < 4

Esempi

i¹⁰⁰ = i⁰ = 1i²²³ = i³ = -ii¹⁰⁷⁰ = i² = -1

100 = 4.25 + 0223 = 4.55 + 31070 = 4.267 + 2

POTENZA DI UN BINOMIO

(a + b)^m = ∑_k=0^m (_mk)a^m-kb^k(_mk) = ^m!/_k!(m-k)!

2^m = (x + 2i)^m = ∑_k=0^m (_mk)x^m-k(2i)^k ; m ∈ ℤ_>0

i^2-m = 1/^2m ; m ∈ ℤ_>0

Esempio

2 = 3 + 2i

2⁵ = (3 + 2i)⁵ = ∑_k=0⁵ (₅k)3^5-k(2i)^k = 0

m ∈ ℤ_>0 = (₅0)3⁵(2i)⁰ + (₅1)3⁴2i¹ + (₅2)3³(2i)² + (₅3)3²(2i)³ +(₅4)3(2i)⁴ + (₅5)(2i)⁵ = 1.243 + 1i + 5.24 - 2i + 10.27(-4) + 10.9

(-8)1 + 5.3 = 16 + 1 - 32i = -597 + 122i

Forma Trigonometrica

Se passo in coordinate polari

• x = ρcosθ
• y = ρsinθ

ρ ∈ [0, +∞)

θ ∉ (-π, π]

z = x + cy = ρcosθ + cρsinθ = ρ (cosθ + i∧nθ)

ρ = √x² + y² = |z|

θ =

• arctan ^y/_x; x > 0
• arctan ^y/_x + π; x < 0, y ≥ 0
• arctan ^y/_x - π; x < 0, y < 0
• π/2; x = 0, y > 0
• -π/2; x = 0; y < 0
• ?

Def

z ≠ 0

Arg(z) = θ ∈ (-π, π]

• argomento principale
• arg(z) = {Arg(z) + 2kπ; k ∈ ℤ}
• argomento

Ogni elemento dell’insieme arg(z) è detto determinòzot dell’argomento di z.

Esempio

Arg(3) = arctan(0) = 0; arg(3) = {2kπ; k ∈ ℤ}

Arg(i) = π/2; arg(i) = {π/2 + 2kπ; k ∈ ℤ}

Arg(-1) = arctan(0) + π = π; arg(-1) = {π + 2kπ; k ∈ ℤ}

Arg (-1 + i) = -arc \tan (-1) + π = -^π/₄ + π = ^3π/₄

arg (-1 + i) = ³/₄ π + 2kπ, k ∈ ℤ

Arg(1/2; -√³/2) = arc \tan(√³) - π = π/₃ - π = -²/₃ π

arg(-1/2; -√³/2) = {-^{2/₃ π + 2kπ, k ∈ ℤ} =}

FORMULE DI DEMOIVRE

z₁ = p₁(cosθ₁ + i \sin θ₁)

z₂ = p₂(cosθ₂ + i \sin θ₂)

z₁z₂ = p₁p₂(cos(θ₁ + θ₂) + i \sin(θ₁ + θ₂))

z₁/_{z2 = p1/p2(cos(θ1 - θ2) + i \sin(θ1 - θ2))}

z₁^m = p₁^m(cos(mθ₁) + i \sin(mθ₁)), m ∈ ℤ \>

Quindi:

arg (z₁z₂) = arg(z₁) + arg(z₂)

arg (z₁/_{z2) = arg(z1) - arg(z2)}

arg ([z₁]^m) = m arg(z₁), m ∈ ℤ \>

ESEMPIO

z₁ = -√3 + i; p₁ = 2; √^√3 + i

z₂ = 1/₃ + √³/₃ i;

arg (z₁) = ⁵/₆ π + kπ; k ∈ ℤ

arg (z₂) = π/₃ + 2kπ; k ∈ ℤ \>

2z₂

arg⁡(z₁z₂)= ⁷⁄₆π + 2kπ, k∈Z{arg(z₁) + arg(z₂)}

Arg(z₁z₂)= ⁵⁄₆π + ⁷⁄₆π = Arg(z₁) + Arg(z₂)

(2) z=-√3 + i = 2 (cos(⁵⁄₆π) + i sin(⁵⁄₆π))

ρ=2

Θ= ⁵⁄₆π

_λ ^{z173 = (-√3 + i)173=2173(cos(173 . 5⁄₆π) + i sin(1735⁄₆π)) =}

= 2¹⁷³ ( cos ^865π⁄₆ + i sin ^865π⁄₆ ) =

= 2 ( cos ^⎧π⁄₆ + i sin ^⎩π⁄₆ ) = 2² ( ^√3⁄₂ + i ¹⁄₂ ) = 2¹⁷² √3 + 2¹⁷² i

875 = 70 . 12 + 1 → ⁸⁶⁵⁄₁₂ = 70 + ¹⁄₁₂ → ³⁶⁵⁄₆ = 70 + 1¹⁄₆

RADICI COMPLESE

z = x + iy = ρ (cosΘ + i sinΘ)

m = ∈ Z > 0

1. ^m√z = { w ∈ &isinfn; : w&sup>m=z}
2. Se w_k = ρ^1⁄m (cos(φ_k) + i sin(φ_k)) ∈&isinfn;
3. Si ha
   1. ρ id √ρ ← radice reale
   2. φ_k = &Theta + 2πk /m, K=0,1,...,m-1
4. se K=m → φ_n = Θ + 2πm - Θ/m = 2π - φ₀

ESEMPIO

√2 = ?

z=2=2(cosΘ+isinΘ) ρ=2 Θ=0

φ₀ = 0 + ⁰⁄₂ = 0

φ₁ = π

w₀ = √2 (cos0 + isin0) = √2

w₁ = √2 (cosπ + isinπ) = -√2

?

p = 2    θ = 0    => r = 3√2

φ₀ = θ + 0 = 0

    (

φ₁ = 0 + 2π/3 = 2π/3 )

φ₂ = 0 + 4π/3 = 4π/3

w₀ = 3√2 (cos0 + isin0) = 3√2

w₁ = 3√2 (cos 2π/3 + isin 2π/3) = 3√2 (-1/2 + √3/2 i)

w₂ = 3√2 (cos 4π/3 + isin 4π/3) = 3√2 (-1/2 - √3/2 i)

p=1; θ = 0

φ₀ = 0

φ₁ = π/2

φ₂ = π

φ₃ = 3/2π

w₀ = 1 (cos0 + isinθ)=1

w₁ = 1 (cosπ/2 + isinπ/2) = i

w₂ = 1 (cosπ + isinπ) = -1

w₃ = 1 (cos3/2π + isin3/2π)= -i

z = -3

p = 3

θ = π

r = √3

φ₀ = π + θ = π/2

φ₁ = π + 2π/2 = 3/2π

w₀ = √3 (cosπ/2 + isinπ/2)=i√3

w₁ = √3 (cos3/2π + isin3/2π)=-i√3