复数z=(1-i)a2-3a+2+i,(a∈R), (1)若z为纯虚数,求z; (2)若复平面内复数z对应的点在第三象限,求a的取值范围.
问题描述:
复数z=(1-i)a2-3a+2+i,(a∈R),
(1)若z为纯虚数,求z;
(2)若复平面内复数z对应的点在第三象限,求a的取值范围.
答
(1)∵z=(1-i)a2-3a+2+i=a2-3a+2+(1-a2)i,(a∈R),则由z为纯虚数可得 a2−3a+2=01−a2≠0,解得a=2,a=1(舍去)∴z=-3i.---------(6分)(2)由题知,a2−3a+2<01−a2<0,解得1<a<2a<−1或a>...