已知复数z1=1+3i,|z2/(z+2i)|=√2,z1*z2为纯虚数,求复数z2

设z2=x+yi
z1*z2=(1+3i)(x+yi)=x-3y+(3x+y)i+为纯虚数,则x=3y
z2=3y+yi |z2|=y√10
|(z+2i)|=2√2
|z2/(z+2i)|=y√10/(2√2)=√2 y=4/√2=2√2
z2 =6√2 +√2 i