|z1|=2,|z2|=3,|z1+z2|=4,则z1/z2=

设z1=2cosa+2sina*i,z2=3cosb+3sinb*i则|z1+z2|=|2cosa+3cosb+i(2sina+3sinb)|=[(2cosa+3cosb)^2+(2sina+3sinb)^2]^(1/2)=[13+12cos(a-b)]^(1/2)=4所以cos(a-b)=1/4z1/z2=(2cosa+2isina)/(3cosb+3isinb)=(2cosa+2isina)(3cosb-3isinb)/9=6cos(a-b)+6isin(a-b)=3/2+6isin(a-b)=3/2 加减 3根号15i/2