1若z1=a+bi(a,b∈R,ab≠0),z2=a-bi,O为坐标原点,复数z对应点Z,则△Z1OZ2的形状

|z1|=√(a^2+b^2)|z2|=√(a^2+b^2)|z1|=|z2|∴△Z1OZ2的形状是等腰三角形(2)a^2+(2+i)a+2xy+(x-y)i=0转化成复数形式得[a^2+2a+2xy]+[a+x-y]i=0复数=0∴实部a^2+2a+2xy=0虚部a+x-y=0a=y-x代入a^2+2a+2xy=0得轨迹方程...