一试11题:由对称性不妨设z1=1,z2=-1/4+isqart3/4,对z=a+bi,令f(z)=asqart3/2+b/2,知zn对应向量必与1,w,w^2对应向量之一同向,对应向量与1,w,w^2对应向量同向的z分别有f(z)=|z|sqart3/2,0,-|z|sqart3/2,而编号相临的zn对应向量不同向,且{|zn|}公比为1/2,表明当m>2,f(z1+...+zm)>sqart3/2-(1+1/4+1/4^2+...)sqart3/8=sqart3/3,从而|z1+...+zm|>=f(z1+...+zm)>sqart3/3.保持z1,z2不变,令z(2k+1)/z2k=z2,z(2k+2)/z(2k+1)=1/4z2,可知∑(z1+...+zm)收敛于1/2+isqart3/6,从而inf|z1+...+zm|=sqart3/3,C=sqart3/3