z= -1 – i√3; |z|=√(1+3)=2; argz=arctg√3-π=-2π/3;
z=2•(cos(-2π/3)+i•sin(-2π/3))= 2•(cos(2π/3)-i•sin(2π/3)).
z =a+bi=Izl(cos(θ )+i sin(θ ))
Пошаговое объяснение:
z= -1 – i√3; |z|=√(1+3)=2; argz=arctg√3-π=-2π/3;
z=2•(cos(-2π/3)+i•sin(-2π/3))= 2•(cos(2π/3)-i•sin(2π/3)).
z =a+bi=Izl(cos(θ )+i sin(θ ))
Пошаговое объяснение: