(1)若z为纯虚数,则
(m−1)(m+2)=0
m−1≠0,
解得m=-2;
(2)若复数z在复平面内对应的点位于第四象限,则
(m−1)(m+2)>0
m−1<0,
解得m<-2;
(3)若m=2,则z=4+i,
a+bi=[4+i+i/4+i−1]=[4+2i/3+i]=
(4+2i)(3−i)
(3+i)(3−i)=[14+2i/10]=[7/5+
1
5i,
∴a=
7
5],b=[1/5],
故a+b=[8/5].
(1)若z为纯虚数,则
(m−1)(m+2)=0
m−1≠0,
解得m=-2;
(2)若复数z在复平面内对应的点位于第四象限,则
(m−1)(m+2)>0
m−1<0,
解得m<-2;
(3)若m=2,则z=4+i,
a+bi=[4+i+i/4+i−1]=[4+2i/3+i]=
(4+2i)(3−i)
(3+i)(3−i)=[14+2i/10]=[7/5+
1
5i,
∴a=
7
5],b=[1/5],
故a+b=[8/5].