设z=a+bi,
z绝对值=2
|z|=√(a^2+b^2)=2,
a^2+b^2=4.(1)
z+3i=a+bi+3i=a+(b+3)i
z+3i绝对值=1
√a^2+(b+3)^2=1
a^2+(b+3)^2=1(2)
由（1）、（2）解之得
a=0
b=-2
故z=-2i