z=a+bi
a=2
√a²+b²=4
a²+b²=16
b²=16-4=12
b=+-√12=+-2√3 cum b<0, b=-2√3=Im(z)
z=2- i*2√3
Pt a il ridica la puterea a 12-a vom folosi scrierea trigonometrica lui z, dand modulul factor comun fortat:
z=4(2/4-i*2√3/4)= 4(1/2-i√3/2)=4(cos5π/3+isin5π/3)
unde α=arctg[ (-√3/2) : (1/2)}= artctg(-√3)= -π/3=(2π-π/3)=5π/3
atunci z^12=[4(cos5π/3+isin5π/3) ]^12= 4^12(cos 60π/3+isin60π/3)=
2 ^24(cos2π+isin2π)=2^24 (1+0*i)=2^24∈R
Re (z^12)=z^12=2^24
x²-4x+16=0
a=1
b=-4
c=16
Δ=16-4*16=-3*16=-48
z1,2= [4+-√(-48)]/2
√-48= √48 * (-1)= √48 * i²= √48*√i²=i*√48=4√3 * i
asadar z1,2=[4+-4√3 *i}/2= 2+-2√3 i
z1= 2+2√3 i
z2=2-2√3 i
