Esercizi di Cauchy, Riemann e residui

Esercizi

1)

β(z) = (_z̅)² Df_β = ϛ (_z̅)² = (_z̅)² = (x-yi)² = x² - y² - 2xyi u(x,y) = x² - y² v(x,y) = -2xy ⟹ (∂u/∂x = 2x) (∂u/∂y = -2y) (∂v/∂x = -2y) (∂v/∂y = -2x) {∂u/∂x + ∂v/∂y = 0 ∂u/∂y + ∂v/∂x = 0} ⟹ H_g = ϛ

2)

_z̅ - 1 = 0 _z̅ = 1, _z = 1, z̅ = 1 Df = ℂ\{1} 1/(x - 4i-1) = 1/(x-1-4i) = ((x-1)+4i)/(x-1+4i) = (x-1 + 4xi)/((x-1)² + 4²) = x-1 μ(x,y) = x-1/((x-1)² + 4²) v(x,y) = 4/((x-1)² + 4²) ⟹ ϛ = ℂ\{1} = Df ∂u(x,y)/∂x = (x-1)² + 4² ∂v(x,y)/∂y = (x-1)² - y²/((x-1)² + 4²)

Esercizi

_f(z) = (^z/_z̅)² D_f = ∅ (^z/_z̅)² = (^{x - yi}/_{x + yi})² U(x, y) = x² - y² V(x, y) = -2xy ∂u/∂x = 2x ∂u/∂y = -2y ∂v/∂x = -2y ∂v/∂y = -2x ∂u/∂x + ∂v/∂y ∂u/∂y + ∂v/∂x ⇒ H_g = ∅

z̅ - 1̅ = 0 z̅ = 1̅ z̅ = 1, z = 1 D_f = ℂ \ {1} ¹/_{x - 4yi - 1} = ¹/_{x - 1 - i} . ^{x - 1 + 4yi}/_{x - 1 + 4yi} = ^{x - 1 + 4yi}/_{(x - 1)² + 4y²} = ^{x - 1}/_{(x - 1)² + 4y²} + ^4y/_{(x - 1)² + 4y²} i

U(x, y) = ^{x - 1}/_{(x - 1)² + 4y²} V(x, y) = ^4y/_{(x - 1)² + 4y²} ⇒ H_g = ℂ \ {1} = D_f ∂U(x, y)/∂x = ^{(x - 1)2 + 4y2}/_{((x - 1)² + 4y²)²} ∂V(x, y)/∂y = ^{-2(x - 1)y}/_{((x - 1)² + 4y²})² ∂V(x, y)/∂y = ^{(x - 1)2 - 4y2}/_{((x - 1)² + 4y²})²

Es.3

∮ Olomorfa tale che μ(x,y) = (e^x + e^-x)x∧y e f'(z).

f(z) = g(x,y) = μ(x,y) + i V(x,y) (∂u(x,y))/∂x) = ∂v(x,y)/∂y ∂u(x,y)/∂y = -∂v(x,y)/∂x ∂u(x,y)/∂x = (e^x - e^-x)∧my

V(x,y) ∫∂u/∂x (x,y)dy == ∫(e^x - e^-x)∧my dy = - (e^x - e^-x)(∧x+y) + C(x) ∂v/∂x = (e^x + e^-x) cos y + C'(x) ∂u(x,y)/∂y = (e^x + e^-x) cos y

(e^x + e^-x) cos y = (e^x + e^-x) cos y - C'(x) C(x) = 0 ⇒ C(x) = K ∈ ℝ

v(x,y) = K - (e^x - e^-x)cos y ; K ∈ ℝ f(z) = (e^x + e^-x) x∧y + i (K - (e^x- e^-x)cos y) f'(z) = ∂g(x,y)/∂x = (e^x - e^-x)∧my - i(e^x+ e^-x) cos y

2)

(x,y) = y² - x² ∫f(z) = ∫(x,y) = u(x,y) + i v(x,y) ∂u(x,y)/∂x = ∂v(x,y)/∂y ∂u(x,y)/∂y = -∂v(x,y)/∂x ∂u/∂x = -2x

V(x,y) = ∫ ∂u/∂x (x,y) dy = ∫ -2x dy = -2xy + C(x) ∂v/∂x = -2y + C'(x) ∂u/∂y = 2y V(x,y) = -2xy + K

f(z) = y² - x² + i ( K - 2xy) ∂f(z)/∂x = ∂f(x,y)/∂x = -2x + i (-2y)

5)

(x,y) = e^1-2x cos(2y) ∂u/∂x = -2 cos(2y) e^1-2x

V(x,y) = ∫ -2 cos(2y) e^1-2x dy = -2 ∫ e^1-2x (cos(2y)) dy 2y = t ⇒ 2 dx = dt ⇒ V(x,y) = e^1-2x sin 2y + C(x)

∂v/∂x = -2 e^1-2x sin 2y + C'(x) ∂u/∂y = + 2 e^1-2x cos(2y) -2e^1-2x sin(2y) = -2e^1-2x sin 2y + C'(x) ⇒ C(x) = K

f(z) = e^1-2x cos(2y) + i ( -e^1-2x sin 2y + K ) ∂f(x)/∂x = ∂f(u,v)/∂x = -2 cos 2y e^1-2x + i (-2 e^1-2x sin 2y)