(1)由题意可知复数z=x2+x-6+(x2-2x-15)i对应的点Z(x2+x-6,x2-2x-15),
落在第三象限内,
∴
x2?2x?15<0
x2+x?6<0,
解得x∈(-3,2).
(2)当Z落在直线x-y-3=0时,
可得x2+x-6-(x2-2x-15)-3=0
解之可得x=2.
(1)由题意可知复数z=x2+x-6+(x2-2x-15)i对应的点Z(x2+x-6,x2-2x-15),
落在第三象限内,
∴
x2?2x?15<0
x2+x?6<0,
解得x∈(-3,2).
(2)当Z落在直线x-y-3=0时,
可得x2+x-6-(x2-2x-15)-3=0
解之可得x=2.