Question

求复数2z^2-1-i轨迹求详细过程的，多谢。z=a+bi,a+b=1

Answer 1

2z^2-1-i

=2(a+bi)^2-1-i

=2(a^2-b^2)-1+(4ab-1)i

=2(a+b)(a-b)-1+(4ab-1)i

=2(2a-1)-1+[4a(1-a)-1]i

=2(2a-1)-1-(2a-1)^2*i

=x+yi (x,y∈R)

∴x=2(2a-1)-1 y=-(2a-1)^2

2a-1=(x+1)/2 y=-[(x+1)/2]^2

所求轨迹方程是y=-(x+1)^2/4