|z|=1,可令z=cost+isint
f(z)=|(1+cost+isint)(cost-isint-i)|=|cost(1+cost)+sint(1+sint)+isintcost-i(1+cost)(1+sint)|
=|1+sint+cost-i(1+sint+cost)|
=|1+sint+cost|*√2
=|1+√2sin(t+π/4)|√2
当t=π/4时,即z=(1+i)/√2时,f(z)取最大值√2+2
此时AZ^2=2+√2
BZ^2=2-√2
AB^2=2
BZ^2+AB^2