设z=x+yi
|z+1+i|^2-|z+-1-i|^2
=(x+1)^2+(y+1)^2-(x-1)^2-(y-1)^2
=x^2+2x+1+y^2+2y+1-x^2+2x-1-y^2+2y-1
=4x+4y
若|z|=1,不妨再设z=cost+isint,x=cost,y=sint
=4(cost+sint)
=4√2sin(t+π/2)∈[-4√2,4√2]
换句话说,只要4a∈[-4√2,4√2],相应地必存在一个t0,使得z0∈M∩N
因此,
a∈[-√2,√2]
有不懂欢迎追问
已知M={Z| Z的模为1,Z是复数},N={Z| (Z+1+i)模的平方减(Z-1-i)模的平方差为4a,Z是复数}