(1)z=x(1+2i)-y(3+i)
=(x-3y)+(2x-y)i
∵z是纯虚数,|z|=√5,y>0
∴x-3y=0,(2x-y)^2=5,y>0
∴x=(3√5)/5,y=√5/5
∴z=√5i
(2)|z(1+i)|=|√5i(1+i)|
=|-√5+√5i|
=√(5+5)=√10