Question

求第三题解答过程，谢谢

Answer 1

解：令f(x)=(cos120°-isin120°)^x (x∈N+)
则f(1)=cos120°-isin120°
f(2)=(cos120°-isin120°)²
=cos²120°-2cos120°×isin120°+i²sin²120°
=-1/2+isin120°=cos120°+isin120°
∴f(4)=(cos120°+isin120°)²
=cos²120°+2cos120°×isin120°+i²sin²120°
=-1/2-isin120°=cos120°-isin120°
即f(1)=f(4)=f(3+1)
∴f(100)=f(3×33+1)=f(1)
故原式=(cos120°+isin120°)(cos120°-isin120°)
=cos²120°-i²sin²120°
=1

注：2π/3=120°
i²=-1
cos120°=-1/2,sin120°=√3/2