因为z-4=(3-2z)i ,所以z-4=3i-2zi那么(1+2i)z=4+3i则z=(4+3i)/(1+2i)=2-i再设w=a+bi,z+w=(a+2)+(b-1)i(z+1)w=(3-i)(a+bi)=(3a+b)+(3b-a)i因为z+w与(z+1)w均为实数所以b-1=0,3b-a=0解得a=3,b=1所以w=3+i又因为w为其一...