Вычислить (1/32)*(-1+sqrt(3)*i)^5

z=x+iy

|z|=√(x²+y²)

cosφ=x/|z|

sinφ=y/|z|

z=-1+√3i

x=-1; y=√3

|z|=√((-1)²+(√3)²)=2

cosφ=x/|z|=-1/2

sinφ=y/|z|=√3/2

φ=2π/3

Тригонометрическая форма:

z=2(cos(2π/3)+isin(2π/3))

z⁵=2⁵(cos5·(2π/3)+isin5·(2π/3))=32(cos(10π/3)+isin(10π/3))=

=32(cos(4π-(2π/3))+isin(4π-(2π/3))=

=32(cos(-(2π/3))+isin(-(2π/3))=

=32(cos(2π/3)-isin(2π/3)=32(1-√3i)

О т в е т. 1-√3i