Resolvendo livro de matemática - Parte 242 скачать в хорошем качестве

Resolvendo livro de matemática - Parte 242

b) z = 2 + 2√3i x = arg(z) p = 4 cos(x) = 1/2 x = 60° 180° 𝝿 60° x 180x = 60𝝿 x = 𝝿/3 sen(x) = √3/2 x = 60° c) z = 4i p = 4 sen(x) = 1 x = 90° x = 𝝿/2 d) z = -2 + 2√3i p = 4 cos(x) = -1/2 x = 120° 180x = 120𝝿 x = 2𝝿/3 9 z = a + bi |z|² + z - z*z' = 3 + 3i |z|² + z - z*z' = 3 + 3i |z| = √(a² + b²) a² + b² + a + bi - (a + bi)(a - bi) = 3 + 3i a² + b² + a + bi - (a² + b²) = 3 + 3i a + bi = 3 + 3i z = 3 + 3i 10 {z ∈ ℂ| |z| = 2} √(a² + b²) = 2 a² + b² = 4 11 i = √-1 z = i5 + 3i4 - 5i³ + 6i² - 3i z = -3 + 3i |z| = 3√2 12 z*z' = 24 (a + bi)(a - bi) = 24 a² + b² = 24 √(a² + b²) = √24 |z| = 2√6 |z| = √(a² + b²) 13 a) zi + 2z' + 1 - i = 0 (a + bi)i + 2(a - bi) + 1 - i = 0 ai - b + 2a - 2bi + 1 - i = 0 2a - b + ai - 2bi = -1 + i 2a - b + (a - 2b)i = -1 + i 2a - b = -1 a - 2b = 1 -4a + 2b = 2 a - 2b = 1 -3a = 3 a = -1 b = -1 z = -1 - i b) z = -1 - i |z| = √2 cos(x) = -√2/2 x = 225° x = 5π/4 14 z1 = 2 - i z2 = x + i |z1*z2|² = 10 (2 - i)(x + i) 2x + 2i - xi + 1 z = 2x + 1 + (2 - x)i |z| = √(2x + 1) + (2 - x)² (2x + 1)² + (2 - x)² = 10 4x² + 4x + 1 + 4 - 4x + x² = 10 x² = 1 x = 1 z = a + bi z = pcos(x) + psen(x)i z = p(cos(x) + sen(x)) Ex1 z = 1 + √3i p = 2 cos(x) = 1/2 x = 60° 180x = 60𝝿 x = π/3 z = 2(cos(π/3) + sen(π/3)i) Ex2 z = 2(cos(π/6) + sen(π/6)i) 𝝿 180° π/6 x x = 30° cos(30°) = a/p a = √3 sen(30°) = b/p b = 1 z = √3 + i 1 a) z = -4√3 - 4i p = 8 cos(x) = -√3/2 sen(x) = -1/2 180x = 210𝝿 x = 7𝝿/6 z = 8(cos(7𝝿/6) + sen(7𝝿/6)i)