Question

已知m∈R,复数z=[m(m-2)]/(m-1)+(m^2+2m-3)i

当m为何值时z对应的点位于复平面第二象限 ？为何值时z对应的点在直线x+y+3=0上？速答谢谢!

Answer 1

z=[m(m-2)]/(m-1)+(m^2+2m-3)i
(1)
z对应的点位于复平面第二象限

那么{[m(m-2)]/(m-1)<0
{ m^2+2m-3>0
==>
{m(m-1)(m-2)<0
{(m+3)(m-1)>0
==>
{1<m<2或m<0
{m<-3或m>1
==>
m<-3或1<m<2
(2)
z对应的点在直线x+y+3=0

那么[m(m-2)]/(m-1)+(m^2+2m-3)+3=0
∴m(m-2)+(m-1)(m²+2m-3)+3(m-1)=0
(m^2+m-3)+(m³+m²-5m+3)=0
m³+2m²-4m=0
m(m²+2m-4)=0
m=0或m=-1-√5或m=-1+√5