Question

calcule o produto 2(cos 2π/3 + i sen 2π/3) por 4(cos π/3+i sen π/3)

Answer 1

Z'•Z'' = -8

Explicação:

Z' = 2(cos 2π/3 + i•sen 2π/3) Ω' = 2

Z'' = 4(cos π/3 + i•sen π/3) Ω'' = 4

Z'•Z'' = Ω'•Ω'' [ cos(2π/3 + π/3) + i•sen(2π/3 + π/3) ]

Z'•Z'' = 2•4 [ cos(π) + i•sen(π) ]

Z'•Z'' = 8 [ -1 + i•0 ]

Z'•Z'' = -8