Analisi matematica 1 - Esercizi complessi

Esercizio 1

z = ^{2 + 5i}⁄_{1 - 3i} = 2 + 5i ; ^{1 + 3i}⁄_{1 - 3i} = ^{2 + 6i + 5i - 15}⁄_{1 + 9} = ^{11i - 13}⁄₁₀

z = -¹³⁄₁₀ + ¹¹⁄₁₀i

Esercizio 2

z = i (5 + 6i)² - ⁴⁄_{1 - i} = i (25 - 36 + 60i) - 4 · ^{(1 + i)}⁄_{(1 + i)} = -11i - 60 - ^{4 + 4i}⁄₂ = -11i - 60 - 2 - 2i = -62 - 13i

Esercizio 3

Z = 1 - i

p = √1 + (-1)² = √2

cosθ = ¹⁄_√2

sinθ = -¹⁄_√2

θ = -^π⁄₄

z = √2 (cos^π⁄₄π + i sen^π⁄₄π)

Esercizio 4

Z = (2 - 2i)⁵

θ = ²⁄_√2 = ^√2⁄₂

sinθ = -²⁄_√2 = -√₂⁄₂

θ = ⁷⁄₄π

Z = 2√2 (cos⁷⁄₄π + i sin^π⁄₄π)

Z = (2√2)⁵ (cos³⁵⁄₂π + i sen³⁵⁄₂π)

Esercizio 1

Z = ^{2 + 5i}/_{1 - 3i} = ^{2 + 5i}/_{-1 + 3i} ⋅ ^{1 + 3i}/_{1 + 3i} = ^{2 + 6i + 5i - 15}/_{1 + 9} = ^{-13 + 11i}/₁₀

Z = ^-13/₁₀ + ¹¹ⁱ/₁₀

Esercizio 2

Z = i(5 + 6i)² - ⁴/_{1 - i} = i(25 - 36 + 60i) - ^{4 ⋅ (1 + i)}/_{1 - i} (1 + i) = -11i - 60 - ^{4 + 4i}/₂ = -11i - 60 - 2 - 2i = -62 - 13i

Esercizio 3

Z = 1 - i

p = √1 + (-1)² = √2

cosθ = ¹/_√2

sinθ = ^-1/_√2

θ = ⁷/₄π

Z = √2 (cos ⁷/₄π + i sin ⁷/₄π)

Esercizio 4

Z = (2 - 2i)⁵

cosθ = ²/_√2 = ^√2/₂

sinθ = ^-2/_√2 = -^√2/₂

θ = ⁷/₄π

X = 2√2 (cos ⁷/₄π + i sin ⁷/₄π)

Z = (2√2)⁵(cos ³⁵/₄π + i sin ³⁵/₄π)√2

Z = 128√2

Z = 128√2 (cos (3π/4) π + i sin (3π/4))e^iπsen

cos

sen

Z = 128 (1 -√2 + i)

Esercizio

Z³ - |Z|² = 0

Z = x + i y

1 ∈ R

√x² + y²

x³ + 3ix² y - 3x y² - i y³ = x² + y²

x³ - 3x y² + i(3x² y - y³) = x² + y²

{x³ - 3x y² = x² + y²

3x² y - y³ = 0

y(3x² - y²) = 0

y = 0

x³ - x² = x² (x - 1) = 0

x = 0   x = 1

3x² y² = y²

3x²

2x² + x² = 0

0 x² (2x + 1) = 0

x³ - 3x³ = x² + 3x² - 8 x³ = 4 x²

x = 0

x = -1/2

y² = 3 ( 1 ) (4)   = 3/4

y = ± √3/2

y₁ = √3/2   y₂ = -√3/2

• [0, 0]
• [1, 0]
• [-1/2, √3/2]
• [√4/2, -√3/2]

Z = 0 , Z = 1

Z = -1/2 + i √3/2 , Z = -1/2 - i √3/2

Secondo metodo

z̅ = ρ(cos θ + i sen θ)

ρ²(cos 3θ + i sen 3θ) = ρ²

ρ³ = ρ²

ρ³ - ρ² = 0

ρ²(ρ - 1) = 0

ρ = 0

ρ = 1

{ cos 3θ = 1

sen 3θ = 0

3θ = 2kπ

3θ = 0

3θ = θ + 2kπ

3θ = 0 + 2kπ

θ = 2/3 kπ

k = 0, 1, 2

θ = 0

θ = 2/3 π

θ = 4/3 π

ρ = 0 : z = 0

ρ = 1

θ = 0 : 1(cos 0 + i sen 0) = 1

z = 1

ρ = 1

θ = 2/3 π : z = 1(cos 2/3 π + i sen 2/3 π) = -1/2 + i√3/2

ρ = 1

θ = 4/3 π : z = -1/2 - i√3/2

Esercizio

z₁ = i

z₂ = 1 - i

z₁ < iz₂ = i

1 + i = i - 1

1 - i = 1 + i

= i - 1

= 1/2 + 1/2 i

i = 1 + i

z₁ = i

z₂ = 1 - i

|z₁| = √(1)² = 1

sinΘ = ¹/_√2 = ¹/_√2

Θ = π/2

|z₂| = √(1)² + (-1)² = √2

cosΘ = √₂/√₂ = ^√2/₂

sinΘ = -1/√₂ = -^√2/₂

Θ = ^π/₄

z₁/_z2 = ¹/_√2 [cos(π/2 - ^π/₄ * π) + i sin(π/2 - ^π/₄ * π)]

z₁/_z2 = ¹/_√2 [cos(-^π/₄ * π) + i sin(-^π/₄ * π)] = ^-1/_√2 [-√₂/2 + i √₂/2] = -¹/₂ + ¹/₂ i

z⁴ = 1

z^a = w

w = 1

|w| = 1

w = 1 [cosΘ + i sinΘ]

Θ = 0 + 2kπ

w = 1 [cos(0 + 2kπ) + i sin(0 + 2kπ)]

z^a = w

z = ^a√w = w¹/_a = ^a√w = ¹/_|1| [cos(2kπ/_a + i sin(2kπ/_a)]

z = [cos(^2kπ/₄) + i sin(π/₄)]

k = 0, 1, 2, 3

• K = 0     z = 1
• K = 1     z = i
• K = 2     z = -1
• K = 3     z = -i