Numeri complessi

z = a + i·b

forma algebrica

z = r·(COS(θ) + i·SIN(θ))

forma trigonometrica

r = √(a^2 + b^2)

z = 2·i - 2·√3

a = - 2·√3

b = 2

r = √((- 2·√3)^2 + 2^2)---> r = 4

TAN(θ) = b/a---> TAN(θ) = 2/(- 2·√3)

TAN(θ) = - √3/3-----> θ = 5·pi/6

quindi:

z = 4·(COS(5/6·pi) + i·SIN(5/6·pi))

-------------------------------------------------

z^n = r^n·(COS(n·θ) + i·SIN(n·θ))

z^12 = 4^12·(COS(12·(5/6·pi)) + i·SIN(12·(5/6·pi)))

z^12 = 16777216

forma algebrica

............................................

i^6 = i³*i³ = (-i)*(-i) =  i² = - 1

z^12 = (2 (i - sqrt(3)))^12 = 2^12(i -sqrt3)^12 = 4096(i^2 -2i*sqrt3 + 3)^6 = 16777216

o anche...

con r =4    e    theta = 5pi/6

z^12 =r^12(cos(12*theta) +i*sin(12*theta)  = 4^12(cos(12(5pi/6)) + i*sin(12(5pi/6))) = 4^12(cos(2(5pi)) + i*sin(10pi)) = 16777216(1 + i*0) = 16777216