03) São dados os números complexos Z1 = 3^(6/6). cos(π/6) + i.sen(π/6) , Z2 = 2^(2/2). cos(π/3) + i.sen(π/3) , Z3 = 4^(3/3). cos(π/4) + i...
calcule:
a) Z1.Z2 R: 6(cos(5π/6) + i.sin(5π/6)) = -3 + 3i√3
b) Z1.Z3 R: 12(cos(π/3) + i.sin(π/3)) = 6 + 6i√3
c) Z3^2 R: 16(cos(π/2) + i.sin(π/2)) = 16i
d) Z2.Z3 R: (cos(5π/4) + i.sin(5π/4)) = -√2
e) Z1^2 R: 9(cos(π/3) + i.sin(π/3)) = 4.5 + 7.794i
f) Z2^3 R: (cos(2π/3) + i.sin(2π/3)) = -1 + i√3
g) Z1.Z2.Z3 R: 24(cos(7π/6) + i.sin(7π/6)) = -6 - 6i√3
a)
b)
c)
d)
e)
f)
g)
Calcular o produto dos complexos
💡 1 Resposta
Ed 
a) Z1.Z2 = (3^(6/6) * 2^(2/2)) * (cos(π/6)cos(π/3) - sen(π/6)sen(π/3) + i[cos(π/6)sen(π/3) + sen(π/6)cos(π/3)]) = 6(cos(5π/6) + i.sin(5π/6)) = -3 + 3i√3 b) Z1.Z3 = (3^(6/6) * 4^(3/3)) * (cos(π/6)cos(π/4) - sen(π/6)sen(π/4) + i[cos(π/6)sen(π/4) + sen(π/6)cos(π/4)]) = 12(cos(π/3) + i.sin(π/3)) = 6 + 6i√3 c) Z3^2 = (4^(3/3))^2 * (cos(π/4)cos(π/4) - sen(π/4)sen(π/4) + i[cos(π/4)sen(π/4) + sen(π/4)cos(π/4)]) = 16(cos(π/2) + i.sin(π/2)) = 16i d) Z2.Z3 = (2^(2/2) * 4^(3/3)) * (cos(π/3)cos(π/4) - sen(π/3)sen(π/4) + i[cos(π/3)sen(π/4) + sen(π/3)cos(π/4)]) = (cos(5π/4) + i.sin(5π/4)) = -√2 e) Z1^2 = (3^(6/6))^2 * (cos(π/6)cos(π/6) - sen(π/6)sen(π/6) + i[cos(π/6)sen(π/6) + sen(π/6)cos(π/6)]) = 9(cos(π/3) + i.sin(π/3)) = 4.5 + 7.794i f) Z2^3 = (2^(2/2))^3 * (cos(π/3)cos(π/3) - sen(π/3)sen(π/3) + i[cos(π/3)sen(π/3) + sen(π/3)cos(π/3)]) = (cos(2π/3) + i.sin(2π/3)) = -1 + i√3 g) Z1.Z2.Z3 = (3^(6/6) * 2^(2/2) * 4^(3/3)) * (cos(π/6)cos(π/3)cos(π/4) - sen(π/6)sen(π/3)sen(π/4) + i[cos(π/6)sen(π/3)sen(π/4) + sen(π/6)cos(π/3)cos(π/4)]) = 24(cos(7π/6) + i.sin(7π/6)) = -6 - 6i√3
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