依题,由复数z=x+yi(x,y∈R),满足│z│=1,得:
x^2+y^2=1
另外:│z-1-i│^2=(x-1)^2+(y-1)^2
=-2(x+y)+3 (注:将x^2+y^2=1带入)
而:1/2=(x^2+y^2)/2 >= [(x+y)/2]^2
所以:(x+y)/2
依题,由复数z=x+yi(x,y∈R),满足│z│=1,得:
x^2+y^2=1
另外:│z-1-i│^2=(x-1)^2+(y-1)^2
=-2(x+y)+3 (注:将x^2+y^2=1带入)
而:1/2=(x^2+y^2)/2 >= [(x+y)/2]^2
所以:(x+y)/2