设α,β为一对共轭复数,若|α-β|=2
|
设α=x+yi,β=x-yi(x,y∈R),
则|α-β|=2|y|=23
∴y=±
.3
∵
=α β2
=x+yi (x-yi)2 x+yi x2-y2-2xyi
=
∈R(x+yi)(x2-y2+2xyi) (x2-y2-2xyi)(x2-y2+2xyi)
∴2x2y+y(x2-y2)=y(3x2-y2)=0
∵y=±3
∴x2=
y2=11 3
∴|α|=
=2x2+y2
故答案为:2
设α,β为一对共轭复数,若|α-β|=2
|
设α=x+yi,β=x-yi(x,y∈R),
则|α-β|=2|y|=23
∴y=±
.3
∵
=α β2
=x+yi (x-yi)2 x+yi x2-y2-2xyi
=
∈R(x+yi)(x2-y2+2xyi) (x2-y2-2xyi)(x2-y2+2xyi)
∴2x2y+y(x2-y2)=y(3x2-y2)=0
∵y=±3
∴x2=
y2=11 3
∴|α|=
=2x2+y2
故答案为:2