Question

|z1|=2,|z2|=3,|z1+z2|=4,则z1/z2=

Answer 1

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)=1/4

z1/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