设z=a+bi,z+2i为实数,a+bi+2i=a,所以b=-2
,z=a-2i,z(1-2i)=(a-2i)(1-2i)=(a-4)-2(a+1)i,为纯虚数,即a-4=0,a=4,
z=4-2i
(1+mi)²=1+2mi-m²=(1-m²)+2mi,在第四象限,
(1-m²)大于0,2m小于0
∣m∣
设z=a+bi,z+2i为实数,a+bi+2i=a,所以b=-2
,z=a-2i,z(1-2i)=(a-2i)(1-2i)=(a-4)-2(a+1)i,为纯虚数,即a-4=0,a=4,
z=4-2i
(1+mi)²=1+2mi-m²=(1-m²)+2mi,在第四象限,
(1-m²)大于0,2m小于0
∣m∣