if z =r(cosθ+isinθ) then z^n=r^n[cos(nθ)+isin(nθ)] and z^(-n)=r^n[cos(-nθ)+isin(-nθ)]
Therefore z^n+z^(-n)=2rcos(nθ) and z^n-z^(-n)=2risin(nθ)
for n=1 and r=1 .........obvious
if z =r(cosθ+isinθ) then z^n=r^n[cos(nθ)+isin(nθ)] and z^(-n)=r^n[cos(-nθ)+isin(-nθ)]
Therefore z^n+z^(-n)=2rcos(nθ) and z^n-z^(-n)=2risin(nθ)
for n=1 and r=1 .........obvious
I don't understand if you meant comment for me or for the OP.
As I said it is true if it is true. Here is the standard proof.