设 z1=cosα+isinα ,z2=2(cosβ+isinβ)
则 z1+z2=(cosα+2cosβ)+i(sinα+2sinβ)=1+√2i
则 cosα+2cosβ=1.(1)
sinα+2sinβ=√2.(2)
两式平方相加得：5+cos(α-β )=3
cos(α-β)=1/2 ,sin(α-β)= ±√3/2
z1/z2= (cosα+isinα )/(2(cosβ+isinβ))
=1/2*(cos(α-β)+isin(α-β))
=1/2*(1/2±√3i/2 )
=1/4±√3i/4