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

由|z2|=1，可假设 z2=cosa+i*sina z2^2 = cos2a + i*sin2a z1 =√3+i = 2 [cos(π/6) + i * sin(π/6) ] z1*z2^2 = 2 * [ cos(π/6+2a) + i * sin(π/6+2a) ] 由z1?z2^2是虚部为负数的纯虚数，得 cos(π/6+2a) = 0，sin(π/6+2a)=-1 可知 π/6+2a = -π/2 得 a = -π&#