设复数z=cosθ+isinθ=e^(iθ),求证cos nθ=Re(z^n)=【z^(2n)+1】/(2z^n)

问题描述:

设复数z=cosθ+isinθ=e^(iθ),求证cos nθ=Re(z^n)=【z^(2n)+1】/(2z^n)
求证cos nθ=Re(z^n)=【z^(2n)+1】/(2z^n)sin nθ=Im(z^n)=【z^(2n)-1】/(2iz^n)
咋来的,求教!

[cosθ+isinθ]^n = cosnθ+isinnθ=> cos nθ=Re(z^n) and sin nθ=Im(z^n)[z^(2n)+1]/(2z^n)=(1/2)z^n + (1/2)z^(-n)=(1/2)[cosnθ+isinnθ] +(1/2)[cosnθ-isinnθ]=cosnθ[z^(2n)-1]/(2iz^n)=[ 1/(2i) ] [ z^n -...