是,用<z>表示z的共軛,exp{z}表示e的z次方,則有
sinz=(exp{iz}-exp{iz})/2i。
令z=x+iy,所以
sin<z> = (exp{i<z>}-exp{i<z>})/2i
= [exp{i(x-iy)}-exp{-i(x-iy)}]/2i
= (exp{y+ix}-exp{-y-ix})/2i
= [e{<y-ix>}-exp{<-y+ix>}]/2i --這裏用到了公式exp{<z>}=<exp{z}>
=<(exp{-y+ix}-exp{y-ix})/2i>
= <[exp{i(x+iy)}-exp{-i(x+iy)}]/2i>
= <sinz>,證畢