1) z是实数,则虚部为0,即m^2 - m - 6 = 0,(m+2)(m-3)=0,m=-2或3
2) z=4+6i,则
m^2-3m = 4
m^2 -m -6 = 6
解得m=4
3) Z对应的点在三象限,说明m^2-3m