1）令z=x+iy, 則
z^2+9/z^2=（x+iy)^2+9/(x+iy)^2為實(shí)數(shù)
=x^2-y^2+2ixy+9(x-iy)^2/(x^2+y^2)^2
=x^2-y^2+2ixy+9(x^2-y^2-2ixy)/(x^2+y^2)^2
因此虛部=0
即2xy-18xy/(x^2+y^2)^2=0
xy[(x^2+y^2)^2-9]=0
所以x=0或y=0,或x^2+y^2=3
軌跡為x,y軸（除去原點(diǎn)）及圓x^2+y^2=3

2) 令z=x+iy
z/(z-1)=(x+iy)/(x-1+iy)=(x+iy)(x-1-iy)/[(x-1)^2+y^2]
=[x(x-1)+y-iy]/[(x-1)^2+y^2]
為純虛數(shù),則有x(x-1)+y=0, 且y≠0
即y=x-x^2,且y≠0, （即x≠0,1)
這是拋物線, 只是除去其中與軸的2個(gè)交點(diǎn)(0,0),(1,0)