Question

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！！！

Answer 1

[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 - z^(-n) ]
=[1/(2i)]( 2isinnθ )
=sinnθ