已知复数z1=√3+i, |z2|=1, z1?z2^2是虚部为负数的纯虚数,求z2

设z2=x+yi，x,y是实数 所以x^2+y^2=1 z1*z2^2=(sqrt(3)+i)*(x^2-y^2+2xyi)=[sqrt(3)(x^2-y^2)-2xy]+[x^2-y^2+2sqrt(3)xy]i 由已知得sqrt(3)(x^2-y^2)-2xy=0 x^2-y^2+2sqrt(3)xy<0 (2) 解得x^2=1/4 y^2=3/4 或x^2=3/4, y^2=1/4 从而x=1/2, y=-